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3 years ago

Can someone help me with this

Mathematics

1 answer:

Varvara68 [4.7K]3 years ago

5 0

Answer:

Step-by-step explanation:

because the point C is 100÷2

and segment diameter meet making it a right angle triangle.

(50 + 90) - 180= 40

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A machine can bind 500 yearbooks in 3 hours. How many hours will it take to bind 1800 yearbooks?

Akimi4 [234]

Answer:

10.8 hours

Step-by-step explanation:

7 0

4 years ago

Find the slope using the formula. (3,-2) and (3, 6)

dangina [55]

Answer:

m = undef

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Algebra I

Slope Formula: $m=\frac{y_2-y_1}{x_2-x_1}$

Step-by-step explanation:

Step 1: Define

Point (3, -2)

Point (3, 6)

Step 2: Find slope m

Substitute: $m=\frac{6+2}{3-3}$
Add/Subtract: $m=\frac{8}{0}$
Divide: m = undef

When we divide anything by zero, we will get undefined. Therefore, this would be a vertical line.

6 0

3 years ago

Si pedro tiene un saldo de -1285 y paga 750. Cual es su nuevo saldo? Con procedimiento por favor

ololo11 [35]

-1285 + 750 = -535 porque el está pagando 750 el balance que le queda es -535

4 0

4 years ago

Read 2 more answers

Multi step equation —20=—4x—6x Answer please

Vikki [24]

Answer:

Solve for x by simplifying both sides of the equation, then isolating the variable

So your Answer is x = 2 Hope this helps :)

Step-by-step explanation:

7 0

4 years ago

The weights of 83 randomly selected windshields were found to have a variance of 1.88. Construct the 95% confidence interval for

Morgarella [4.7K]

Answer:

95% confidence interval for the population variance = (1.42 , 2.62).

Step-by-step explanation:

We are given that the weights of 83 randomly selected windshields were found to have a variance of 1.88.

So, firstly the pivotal quantity for 95% confidence interval for the population variance is given by;

P.Q. = $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, $s^{2}$ = sample variance = 1.88

$\sigma^{2}$ = population variance

n = sample of windshields = 83

So, 95% confidence interval for population variance, $\sigma^{2}$ is;

P(58.85 < $\chi^{2} __8_2$ < 108.9) = 0.95 {As the table of $\chi^{2}$ at 82 degree of freedom

gives critical values of 58.85 & 108.9}

P(58.85 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 108.9) = 0.95

P( $\frac{ 58.85}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{108.9}{(n-1)s^{2} }$ ) = 0.95

P( $\frac{ (n-1)s^{2}}{108.9 }$ < $\sigma^{2}$ < $\frac{ (n-1)s^{2}}{58.85 }$ ) = 0.95

95% confidence interval for $\sigma^{2}$ = ( $\frac{ (n-1)s^{2}}{108.9 }$ , $\frac{ (n-1)s^{2}}{58.85 }$ )

= ( $\frac{ (83-1)\times 1.88}{108.9 }$ , $\frac{ (83-1)\times 1.88}{58.85 }$ )

= (1.42 , 2.62)

Therefore, 95% confidence interval for the population variance of the weights of all windshields in this factory is (1.42 , 2.62).

8 0

3 years ago