Ms. fudge recently gave an algebra 1 test. the scores were normally distributed with a mean of 74 and a standard deviation of 8. find each of the following. write as a percent.

Answer:

Step-by-step explanation:

The given information is not sufficient to determine each angle measure since the quadrilateral is irregular.

It would be good to see the picture at least.

Possible options are A or B but we can't confirm without additional detals, option C can't be a correct choice as 180° represents a straight line.

Answer:

The required angles are 15, 75 and 90

Step-by-step explanation:

1x + 5x+ 6x = 180
or 12x = 180
180 ÷ 12 = 15
hence,
15, 75, 90.

Answer:

2,3

Step-by-step explanation:

you count up 2,then count over three

Answer:  Distributive property  a(b+c) = ab + ac

-5(7)x + 7(7) = -35x + 49 (D)

Answer:

Step-by-step explanation:

Simply put, it is similar to finding the area of a circle.

Therefore, a=pi*r^2

a=pi*23^2

a=529pi

a=1661.9 ft^2

Answer:

4√3

Step-by-step explanation:

Hello!

This is a 30° 60° 90° because it has a 90° angle, a 30° angle, and the missing angle is 60° (sum of angles in a triangle is 180°). This triangle has special properties.

In a 30°-60°-90° triangle, the leg opposite 30° is always half the hypotenuse.

This means that the hypotenuse would be 2 times 4, or 8.

The leg opposite to 60° in a right triangle of this sort is the length of a leg multiplied by √3.

So the measure of x is 4√3.

An image is attached for your reference.

The value of side labeled x using Trigonometric ratio is  [tex]{4}{\sqrt 3}[/tex].unit.

Given:

Angle = 30

Perpendicular or opposite side= 4

Using Trigonometric ratio

tan 30 = Opposite side / Adjacent side

Substituting Opposite side= 4 and Adjacent side = 4 in above ratio

So, tan 30 = 4 / x

As, the value of tan 30 = [tex]\dfrac{1}{\sqrt 3}[/tex] then

tan 30 = 4 / x

[tex]\dfrac{1}{\sqrt 3}[/tex]  = [tex]\dfrac{4}{x}[/tex]

solving for x

x = [tex]{4}{\sqrt 3}[/tex]

Thus, the value of x is [tex]{4}{\sqrt 3}[/tex].

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➝ Hypotenuse of triangle ( a ) = 21.63 mm

➝ Hypotenuse of triangle ( b ) = 150 mm

➝ Hypotenuse of triangle ( c ) = 111.80 mm

[tex] \quad\rule{300pt}{1.5pt}\quad[/tex]

We have to find the length of hypotenuse in the given 3 triangles, which can be done by using Pythagoras theorem.

" In a right angled triangle, the square of hypotenuse side is equal to the sum of square of other two sides "

[tex] \qquad \bull \:{\pmb{\mathfrak{ h^2 = b^2 + p^2}}}[/tex]

And we have to convert the answer to the units indicated in red i.e, in mm.

Since 1cm = 10 mm, we will convert the given values of length of side in mm before putting the values in the formula

[tex] :\implies\qquad \sf{ h^2 = b^2 + p^2}[/tex]

[tex] :\implies\qquad \sf{ h =\sqrt{b^2 + p^2}}[/tex]

[tex] :\implies\qquad \sf{h=\sqrt{ (12)^2 + (18)^2 }}[/tex]

[tex] :\implies\qquad \sf{ h= \sqrt{144 + 324}}[/tex]

[tex] :\implies\qquad \sf{ h = \sqrt{468}}[/tex]

[tex] :\implies\qquad\underline{\underline{\pmb{ \sf{ h = 21.63 mm}}}}[/tex]

[tex] :\implies\qquad \sf{ h^2 = b^2 + p^2}[/tex]

[tex] :\implies\qquad \sf{h =\sqrt{b^2 + p^2} }[/tex]

[tex] :\implies\qquad \sf{ h = \sqrt{(90)^2+(120)^2}}[/tex]

[tex] :\implies\qquad \sf{ h=\sqrt{8100+14400}}[/tex]

[tex] :\implies\qquad \sf{ h =\sqrt{22500}}[/tex]

[tex] :\implies\qquad\underline{\underline{\pmb{ \sf{h = 150mm}}} }[/tex]

[tex] :\implies\qquad \sf{h^2 = b^2 + p^2 }[/tex]

[tex] :\implies\qquad \sf{ h=\sqrt{b^2 + p^2}}[/tex]

[tex] :\implies\qquad \sf{ h =\sqrt{(100)^2)+(50)^2}}[/tex]

[tex] :\implies\qquad \sf{ h=\sqrt{10000+2500}}[/tex]

[tex] :\implies\qquad \sf{ h =\sqrt{12500}}[/tex]

[tex] :\implies\qquad \underline{\underline{\pmb{\sf{h = 111.80mm}}} }[/tex]

Hypotenuse of triangle ( a ) = 21.63 mm

Hypotenuse of triangle ( b ) = 150 mm

Hypotenuse of triangle ( c ) = 111.80 mm

find the length of hypotenuse in the given 3 triangles, which can be done by using Pythagoras theorem.

[tex]h^2 = b^2 + p^2[/tex]

And we have to convert the answer to the units indicated in red i.e, in mm.

Since 1cm = 10 mm, we will convert the given values of length of side in mm before putting the values in the formula

[tex]For \: \: triangle ( a )

\qquad \sf{ h^2 = b^2 + p^2}[/tex]

[tex]\qquad \sf{ h =\sqrt{b^2 + p^2}}[/tex][tex]\qquad \sf{h=\sqrt{ (12)^2 + (18)^2 }}[/tex][tex]\qquad\sf{h=\sqrt{ (12)^2 + (18)^2 }} \\ \\ \qquad \sf{ h= \sqrt{144 + 324}} \\ \\ \qquad \sf{ h = \sqrt{468}}

\\ \\\qquad\underline{\underline{\pmb{ \sf{ h = 21.63 mm}}}} \\ \\ For \: \: triangle ( b ) \qquad \sf{ h^2 = b^2 + p^2} \\ \\ \qquad \sf{h =\sqrt{b^2 + p^2} } \\ \\ \qquad \sf{ h = \sqrt{(90)^2+(120)^2}} \\ \\ \qquad \sf{ h=\sqrt{8100+14400}} \\ \\

\qquad \sf{ h =\sqrt{22500}} \\ \\\qquad\underline{\underline{\pmb{ \sf{h = 150mm}}} } \\ \\ For \: \: triangle ( c ) \qquad \sf{h^2 = b^2 + p^2 } \\ \\ \qquad \sf{ h=\sqrt{b^2 + p^2}} \\ \\\qquad\sf{ h=\sqrt{(100)^2)+(50)^2}} \\ \\\qquad\sf{ h=\sqrt{10000+2500}} \\ \\ \qquad \sf{ h =\sqrt{12500}} \\ \\

\qquad\underline{\underline{\pmb{\sf{h = 111.80mm}}} } [/tex]

Answer:

D

Step-by-step explanation:

1/2 hour -> 1/3 pool

multiply this by 2

1 hour -> 2/3 pool

so D 2/3 pool

The fraction of the pool that would be filled per hour at the given rate is: D. 2/3 of the pool.

A constant rate can be described as a quantity that changes steadlity over time.

The rate that the pool gets filled is given as:

1/2 hr = 1/3 of the water that would be filled in the pool

1 hr would be: 1/3 × 2 = 2/3

Therefore, the fraction of the pool that would be filled per hour at the given rate is: D. 2/3 of the pool.

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Answer:

Ur mom

Step-by-step explanation:

Ur mom JK here's the answer 11:45 A.M. because 15 + 15 + 15 = 45 15 is a quarter so yeah Bois is 11:45 A.M Good luck

Answer:

Step-by-step explanation:

Given that area of an equilateral traingle 50 square units.

Let the length of side of equilateral traingle be 'x' units. To calculate length of each side we will use the formula of area of equilateral traingle:

[tex] \: \: { \underline { \boxed{\pmb { \sf{ \purple {Area_{(equilateral \: traingle)} = \dfrac{\sqrt 3}{4} \times (Side)^2}} }}}} \\ [/tex]

By substituting the values in above formula:

[tex]~[/tex]

[tex] : \implies \sf \: \: 50 = \dfrac{ \sqrt{3} }{4} \times {(x)}^{2} \\ \\ \\ : \implies \sf \: \: {(x)}^{2} = \dfrac{50 \times 4}{ \sqrt{3} } \\ \\ \\ : \implies \sf \: \: {(x)}^{2} = \dfrac{200}{ \sqrt{3} } \\ \\ \\ : \implies \sf \: \: x = \sqrt{ \dfrac{200}{ \sqrt{3} } } \\ \\ \\ : \implies \: \: { \boxed{\pmb{ \frak{ x = 10.75 \: units }}}}\\ \\ \\ [/tex]

Answer:

each pet bird got 1/8 of the bag of birdseed

Step-by-step explanation:

the easiest way to do this is to divide the bag into eighths. if you divide the bag into eighths, 4/8 of the bag, being half, goes into the bird feeder. the other 4/8 of the bag is divided among Mr. Wilson's four pet birds, leaving each bird to get 1/4 of the bag when split evenly.

hope this helps

Answer:

  N = 74(1/2)^(t/2.8)

Step-by-step explanation:

The exponential function expressing a half-life relation can be written ...

  amount = (initial amount) × (1/2)^(t/(half-life))

For the numbers given in this problem, this is ...

  N = 74(1/2)^(t/2.8)

__

Some folks like to express these relations in the form ...

  N  = 74e^(-kt)

In this form, the value of k is ...

  k = ln(2)/(half-life) ≈ 0.693147/2.8 ≈ 0.24755

  N = 74e^(-0.24755t)

Perpendicular lines have slopes that are opposite reciprocals.

This means we take a number, flop it over, and change its sign.

Here we have the number:-

[tex]\boxed{\frac{7}{3}}[/tex]

Flop it over:-

[tex]\boxed{\frac{3}{7} }\longleftarrow\sf{the~numerator~and~denominator~switch~places}[/tex]

Change its sign:-

[tex]\boxed{-\frac{3}{7}}[/tex]

Hence, the slope of line e is:-

[tex]\bigstar{\boxed{\pmb{Slope=-\frac{3}{7}}}}[/tex]

Hope everything is clear; if you need any explanation/clarification, kindly let me know, and I will comment and/or edit my answer :)

Answer:

[tex] \frac{ \alpha + \beta + \gamma }{ - d} [/tex]

Step-by-step explanation:

If we simplify that fraction, we get

[tex] \frac{ \alpha + \beta + \gamma }{ \alpha \beta \gamma } [/tex]

Keep that in mind.

If y, a ,b are zeroes of the cubic polynomial, then that means

[tex](x - \alpha )(x - \beta )(x - \gamma )[/tex]

make up the polynomial.

Notice that leading xoeffeicent will be 1, so the roots will multiply to

[tex] - d[/tex]

so

[tex] \alpha \beta \gamma = - d[/tex]

which gives us

[tex] \frac{ \alpha + \beta + \gamma }{ - d} [/tex]

Proof:

Consider the function

[tex](x - 2)(x - 3)(x - 5)[/tex]

The roots are 2, 3, 5.

D is -30 so we get

Using the value,

[tex] \frac{2 + 3 + 5}{ 30} = \frac{1}{3} [/tex]

If we use the orginal equation, we get

[tex] \frac{1}{6} + \frac{1}{10} + \frac{1}{15} = \frac{10}{30} = \frac{1}{3} [/tex]

Answer:

Hey,mate

Notice that leading xoeffeicent will be 1, so the roots will multiply to

The roots are 2, 3, 5.

[tex]\sqrt{2} \sqrt{3} \sqrt{5}[/tex]

D is -30

Answer:

8/3 < 8/2

D

Step-by-step explanation:

8/3 = 2.67

8/2 = 4

So, when fractions have equivalent (same) numerators, the one with the lesser denominator is the greater fraction.

There is more left if 8 is divided by 2 than when it is divided by 3

[Answer]8/3 < 8/2

[Answer] D

PLEASE RATE!! I hope this helps!!

If you have any questions comment below

Answer:

c 7-x

Step-by-step explanation:

Answer:

line

Step-by-step explanation:

A line is a set of points that extends infinitely in both directions.

The function f(x) = x² and g(x) = |x|² are the same since they both give a positive value.

From the complete information, it should be noted that the function f(x) = x² and g(x) = |x|² are the same since they both give a positive value.

Also, f(x) = x³ and g(x) = |x|³ are not equal. This is because both functions are not equal for x < 0. This can be illustrated when x = -1. In this case, f(x) = -1 and g(x) = 1.

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The function f(x) = x² and g(x) = |x|² are the same since they both give a positive value.

Two functions are equal if they have the same domain and codomain and their values are the same for all elements of the domain.

The function f(x) = x² and g(x) = |x|² are the same since they both give a positive value.

Also, f(x) = x³ and g(x) = |x|³ are not equal.

This is because both functions are not equal for x < 0.

This can be illustrated when x = -1. In this case, f(x) = -1 and g(x) = 1.

Hence, the function f(x) = x² and g(x) = |x|² are the same since they both give a positive value.

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Using an exponential function, it is found that:

a) [tex]N(t) = 75(0.5)^{\frac{t}{3.8}}[/tex]

b) 37.5 grams of the gas remains after 3.8 days.

c) The amount remaining will be of 10 grams after approximately 11 days.

A decaying exponential function is modeled by:

[tex]A(t) = A(0)(1 - r)^t[/tex]

In which:

Item a:

We start with 75 grams, and then work with a half-life of 3.8 days, hence the amount after t daus is given by:

[tex]N(t) = 75(0.5)^{\frac{t}{3.8}}[/tex]

Item b:

This is N when t = 3.8, hence:

[tex]N(t) = 75(0.5)^{\frac{3.8}{3.8}} = 37.5[/tex]

37.5 grams of the gas remains after 3.8 days.

Item c:

This is t for which N(t) = 10, hence:

[tex]N(t) = 75(0.5)^{\frac{t}{3.8}}[/tex]

[tex]10 = 75(0.5)^{\frac{t}{3.8}}[/tex]

[tex](0.5)^{\frac{t}{3.8}} = \frac{10}{75}[/tex]

[tex]\log{(0.5)^{\frac{t}{3.8}}} = \log{\frac{10}{75}}[/tex]

[tex]\frac{t}{3.8}\log{0.5} = \log{\frac{10}{75}}[/tex]

[tex]t = 3.8\frac{\log{\frac{10}{75}}}{\log{0.5}}[/tex]

[tex]t \approx 11[/tex]

The amount remaining will be of 10 grams after approximately 11 days.

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5/6 x 4/9 = 10/27 as a fraction form. The work for 5/6 times 4/9 as a fraction provides more insight of how to find what is 5/6 times 4/9 in fraction form and the different variations of such problems.

Answer:

the side s is 18

Step-by-step explanation:

s=?

Area of square(A)=324

Now,

A=s²

324=s²

√(324)=s

s=18

Answer:

around 8 or 9 years

Step-by-step explanation:

Answer:

The angle of KLM is 50°

Step-by-step explanation:

Used protractor

Answer:

∠ KLM ≈ 33.7°

Step-by-step explanation:

using Pythagoras' identity in right triangle JKM

KM²  = JK² + JM² = 7² + 4.5² = 49 + 20.25 = 69.25 ( take square root of both sides )

KM = [tex]\sqrt{69.25}[/tex] ≈ 8.322 cm

using the sine ratio in right triangle KLM

sin KLM = [tex]\frac{opposite}{hypotenuse}[/tex] = [tex]\frac{KM}{LM}[/tex] = [tex]\frac{8.322}{15}[/tex] , then

∠ KLM = [tex]sin^{-1}[/tex] ( [tex]\frac{8.322}{15}[/tex] ) ≈ 33.7° ( to 3 significant figures )

Answer:

29.5

Step-by-step explanation:

14sq + 26sq = 872

sq root of 872 is 29.5

Step-by-step explanation:

Table A represents an arithmetic sequence with a common difference d=+7.5

so for x=4, it's y³+7.5=25+7.5=32.5

for x=5, it's y⁴+7.5=32.5+7.5=40

Table B represents a geometric sequence with a common ratio r=4, so you're essentially multiplying by 4.

for x=4, y⁴=y³×4=160×4=640

for x=5, y⁵=y⁴×4=640×4=2560

Answer:

1

Step-by-step explanation:

If you're given 3 side lengths and 3 measurements of the angles , there's only one triangle that you may draw. You also can draw distinctive  triangles given 3 aspect lengths or  aspect lengths and the blanketed distinctive measure.

Answer:

1

Step-by-step explanation:

if you are given 3 measurements you are kind of stuck making that one triangle, no matter which way you shape it you have to have those three angles so you can't make more than that one triangle

Answer:

63

Step-by-step explanation:

Multiply each side by 9

4y=252

Divide by 4

y=63

Answer:

y= 63

Step-by-step explanation:

Answer: 75, 40, 65

Step-by-step explanation:

Let the angles measure 15x, 8x, and 13x.

Since the sum of the angles of a triangle is 180 degrees,

15x+8x+13x=180, meaning that x=5.

So, the angles are 15(5)=75 degrees, 8(5)=40 degrees, and 13(5)=65 degrees.