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4 years ago

15

angle 1 and angle 2 are complementary angles. if the measure of angle 1 is twice the measure of angle 2, find the measure of ang

le 1 and angle 2

1 answer:

ikadub [295]4 years ago

8 0

Angle 1 is 60 degrees and Angle 2 is 30 degrees.

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I suck at area of a circle

IgorC [24]

Answer:

See below.

Step-by-step explanation:

Formula: A = πr²

Since the radius is 4, we fill in the formula with the radius.

A = (3.14)4²

A = (3.14)16

A = 50.24

-hope it helps

7 0

2 years ago

Tracy has 7/8 pounds of butter in her refrigirator. She uses 2/3 of it for cookie batter. How much of the butter did she use?

adelina 88 [10]

Answer:

16/24

Step-by-step explanation:

So you need to get a common denominator before you try to answer the whole question. Turn the denominator into 24. Since you would multiply 8 by 3, multiply the 7 by 3. That would make 21/24. Make the 3 a 24, multiplying it by 8 and doing the same to the 2 making 16/24.

5 0

3 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.

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

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

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

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

= (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

Mary Beth has $1500 in a savings account on January 1. She wants to have at least $800 by May 1. She estimates that she can wit

DerKrebs [107]

1500-38.5x > 800
x= Number of weeks

5 0

3 years ago

PLEASE HELP THIS IS REALLY IMPORTANT

Svet_ta [14]

Answer:

a) $x^2-x-20\\$

b) what?

c) -70

Step-by-step explanation:

If you have any questions feel free to ask in the comments - Mark

sorry for the terrible answer

3 0

3 years ago