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evablogger [386]

4 years ago

14

Mr. Raysor bought crayons and markers for his school. He bought 5 times as many crayons as markers.He bought 30 boxes a total of

30 boxes. How many boxes of crayons and bixes of markers did Mr. Raysor buy?
complete the bar model. Write an equation and solve

1 answer:

adoni [48]4 years ago

7 0

X=markers

5x= crayons

5x+x= 30

6x=30

X=5

5 markers. 25 crayons

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Solve the triangle. Round your answers to the nearest tenth.

mixas84 [53]

Answer:

Angles: $\angle A = 43^{\circ}$ $\angle B = 55^{\circ}$ $\angle C = 82^{\circ}$

Sides: $BC = 20$ $AB = 29$ $AC =24$

Step-by-step explanation:

See attachment for complete question

From the attachment, we have:

$AB = 29$

$AC =24$

$\angle C = 82^{\circ}$

Required

Complete the missing side and missing angles

To calculate angle B, we apply sine laws:

$\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}$

In this case:

$\frac{AB}{sinC}=\frac{AC}{sinB}$

This gives:

$\frac{29}{sin(82^{\circ})}=\frac{24}{sinB}$

$\frac{29}{0.9903}=\frac{24}{sinB}$

Cross Multiply

$sinB * 29 = 24 * 0.9903$

Divide both sides by 29

$sinB = \frac{24 * 0.9903}{29}$

$sinB = 0.8196$

Take arcsin of both sides

$B = sin^{-1}(0.8196)$

$B = 55^{\circ}$

So:

$\angle B = 55^{\circ}$

To solve for the third angle, we make use of:

$\angle A + \angle B + \angle C = 180^{\circ}$

This gives:

$\angle A + 55^{\circ} + 82^{\circ} = 180^{\circ}$

$\angle A + 137^{\circ}= 180^{\circ}$

$\angle A = 180^{\circ}- 137^{\circ}$

$\angle A = 43^{\circ}$

Hence, the angles are:

$\angle A = 43^{\circ}$ $\angle B = 55^{\circ}$ $\angle C = 82^{\circ}$

To calculate the length of the third side, we apply cosine law

$BC^2 = AB^2 + AC^2 - 2*AB*AC*cosA$

$BC^2 = 29^2 + 24^2 - 2*29*24*cos(43^{\circ})$

$BC^2 = 841+ 576 - 1392*cos(43^{\circ})$

$BC^2 = 841+ 576 - 1392*0.7314$

$BC^2 = 841+ 576 - 1018.11$

$BC^2 = 398.89$

Take the square root of both sides

$BC = \sqrt{398.89$

$BC = 19.9722307217$

$BC = 20$

5 0

3 years ago

The Han dynasty lasted from 206 B.C. to 9 A.D. How many years did the dynasty last?

cricket20 [7]

Answer:

The Han dynasty last was, 215 years

Step-by-step explanation:

Given statement: The Han dynasty lasted from 206 B.C. to 9 A.D.

To find how many years did the dynasty last.

B.C. represents Before Christ

A.D. represents After Death

To find the difference between two dates,

When the endings are the same you subtract

When they are different you add

You can see that the endings in the given 206 B.C. and 9 A.D. are B.C. and A.D. are different.

Just add these 206 and 9 we get;

$206 + 9 = 215$ years

Therefore, 215 years did the Han dynasty last.

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3 years ago

A cylinder has a diameter of 6 ft and height of 9 ft. What is the volume of this cylinder?

agasfer [191]

The answer is 254.47 ft

Step-by-step explanation:

V=pi×r^2×h

pi=3.14

r=d/2

r=6/2=3

v=(3.14)×3^2×9

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4 years ago

Verify that the functions are probability mass functions, and determine the requested probabilities.

Free_Kalibri [48]

Answer:

a) 3/64 = 0.046 (4.6%)

b) 63/64 = 0.9843 (98.43%)

c) 1/64 = 0.015 (1.5%)

d) 1/4 = 0.25 (25%)

Step-by-step explanation:

in order to verify that the f(x) is a probability mass function , then it should comply the requirement that the sum of probabilities over the entire space of x is equal to 1. Then

∑f(x)*Δx = 1

if f(x)=(3/4)(1/4)^x , x = 0, 1, 2, ...

then Δx=1 and

∑f(x) = (3/4)∑(1/4)^x = (3/4)* [ 1/(1-1/4)] = (3/4)*(4/3) = 1

then f represents a probability mass function

a) P(X = 2)= f(x=2) = (3/4)(1/4)^2 = 3/64 = 0.046 (4.6%)

b) P(X ≤ 2) = ∑f(x) = f(x=0)+ f(x=1) + f(x=2) = (3/4) + (3/4)(1/4) + 3/64 = 63/64 = 0.9843 (98.43%)

c) P(X > 2)= 1- P(X ≤ 2) = 1 - 63/64 = 1/64 = 0.015 (1.5%)

d) P(X ≥ 1) = 1 - P(X < 1) = 1 - f(x=0) = 1- 3/4 = 1/4 = 0.25 (25%)

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3 years ago

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loris [4]

Answer:

The answer is option (3) Multicollinearity

Kindly find an attached copy of a diagram to the given question

Step-by-step explanation:

Solution

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Multicollinearity: Refers to a method where variables that are independent in a regression model are associated.

This correlation or association of variables tends to be a problem, because variables that are independent should remain independent.

3 0

3 years ago