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2 years ago

Which of these is equal to cos 19°?

1 answer:

4 0

Hello there.

Question: <span>Which of these is equal to cos 19°?

Answer: There aren't any options so i can only give the value for it which is 0.9455

</span>Hope This Helps You!
Good Luck Studying ^-^

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Sherman has 3 cats and 2 dogs. he wants to buy a toy for each of his pets. Sherman has 22$ to spend on pet toys. How much can he

WARRIOR [948]

Sherman wants to spend on each pet, so we are assuming he wants to spend the same amount. He has 5 pets in all (3 cats, 2 dogs)

Divide 22 with 5

22/5 = 4.4

He can spend $4.40, or in fraction form:

4 4/10 (Mixed fraction)

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Micayla wants the print shop to reduce the size of her painting but keep the ratio of length to width so that it will fit her fr

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Answer:

The width of the frame must be 12 inches (15in by 12in)

Step-by-step explanation:

Given

See attachment for painting and frame

Required

Determine the dimension of the frame

To do this, we make use of the following scale ratio.

$Scale = Length : Width$

For the painting,

$Scale = 20: 16$

For the frame,

$Scale = 15: w$

Since she wants to maintain the same ratio; both scale must be the same.

i.e.

$15 : w = 20 : 16$

Express as fraction

$\frac{w}{15} = \frac{16}{20}$

$\frac{w}{15} = 0.8$

Make w the subject

$w = 15 * 0.8$

$w = 12$

<em>The width of the frame must be 12 inches i.e. 15in by 12in frame</em>

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3 years ago

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3 years ago

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Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in

Law Incorporation [45]

Answer:

(a) 900

(b) [567.35 , 1689.72]

(c) [23.82 , 41.11]

Step-by-step explanation:

We are given that a sample of 20 days of operation shows a sample mean of 290 rooms occupied per day and a sample standard deviation of 30 rooms i.e.;

Sample mean, $xbar$ = 290 Sample standard deviation, s = 30 and Sample size, n = 20

(a) Point estimate of the population variance is equal to sample variance, which is the square of Sample standard deviation ;

$\sigma^{2}$ = $s^{2}$ = $30^{2}$

$\sigma^{2}$ = 900

(b) 90% confidence interval estimate of the population variance is given by the pivotal quantity of $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

P(10.12 < $\chi^{2}__1_9$ < 30.14) = 0.90 {At 10% significance level chi square has critical

values of 10.12 and 30.14 at 19 degree of freedom}

P(10.12 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 30.14) = 0.90

P( $\frac{10.12}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{30.14}{(n-1)s^{2} }$ ) = 0.90

P( $\frac{(n-1)s^{2} }{30.14}$ < $\sigma^{2}$ < $\frac{(n-1)s^{2} }{10.12}$ ) = 0.90

90% confidence interval for $\sigma^{2}$ = $[\frac{19s^{2} }{30.14} , \frac{19s^{2} }{10.12}]$

= $[\frac{19*900 }{30.14} , \frac{19*900 }{10.12}]$

= [567.35 , 1689.72]

Therefore, 90% confidence interval estimate of the population variance is [567.35 , 1689.72] .

P( $\sqrt{\frac{(n-1)s^{2} }{30.14}}$ < $\sigma$ < $\sqrt{\frac{(n-1)s^{2} }{10.12}}$ ) = 0.90

90% confidence interval for $\sigma$ = $[\sqrt{\frac{19s^{2} }{30.14}} , \sqrt{\frac{19s^{2} }{10.12}} ]$

= [23.82 , 41.11]

Therefore, 90% confidence interval estimate of the population standard deviation is [23.82 , 41.11] .

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3 years ago