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4 years ago

Find tan B. AB=17,BC=15 AC= 8

Mathematics

1 answer:

Aneli [31]4 years ago

5 0

Answer:

a is the ans;8/15

Step-by-step explanation:

we have tanb = p/ b

so,by taking reference angle b we get,

p=8 and b= 15.

therefore; tanb=8/15

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5(3x + 5) ( Please Help )

BARSIC [14]

A(b+c) = ab + ac

5(3x + 5) = 15x + 25

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3 years ago

A circle has a diameter of 2.5 ft.

alukav5142 [94]

A = pi * r^2
D = 2.5 so r = 1.25
A = 3.14 * 1.25^2
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8 0

3 years ago

A vial holds 2 milliliters of pixie dust. How many vials can we fill with 13 milliliters of pixie dust?

Bezzdna [24]

Answer: 6 and a half

Step-by-step explanation: 13 is an uneven number so you divide 13 by 2 and get 6.5

3 0

3 years ago

You measure 50 textbooks' weights, and find they have a mean weight of 37 ounces. Assume the population standard deviation is 5.

Hoochie [10]

Answer:

90% confidence interval for the true population mean textbook weight is [35.79 ounces , 38.21 ounces].

Step-by-step explanation:

We are given that you measure 50 textbooks' weights, and find they have a mean weight of 37 ounces.

Assume the population standard deviation is 5.2 ounces.

Firstly, the Pivotal quantity for 90% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean weight = 37 ounces

$\sigma$ = population standard deviation = 5.2 ounces

n = sample of textbooks = 50

$\mu$ = true population mean textbook weight

Here for constructing 90% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 90% confidence interval for the population mean, $\mu$ is ;

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5%

level of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

P( $\bar X -1.645 \times {\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X +1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.90

90% confidence interval for $\mu$ = [ $\bar X -1.645 \times {\frac{\sigma}{\sqrt{n} } }$ , $\bar X +1.645 \times {\frac{\sigma}{\sqrt{n} } }$ ]

= [ $37-1.645 \times {\frac{5.2}{\sqrt{50} } }$ , $37+1.645 \times {\frac{5.2}{\sqrt{50} } }$ ]

= [35.79 , 38.21]

Therefore, 90% confidence interval for the true population mean textbook weight is [35.79 ounces , 38.21 ounces].

7 0

3 years ago

PLEASE HELP ASAP!!!!!!

lorasvet [3.4K]

I'm not sure what the question is but it looks like it's B

5 0

3 years ago