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3 years ago

PLEASE HURRRYY!!! I need help

Mathematics

1 answer:

VashaNatasha [74]3 years ago

3 0

Answer:

C) 8 ft

Step-by-step explanation:

you use the pythagorean theorem to find the height.

a² + b² = c²

a = radius

b = height

c = slant height

15² + h² = 17²

h = 8

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What is 60= 3/4h i am doing BIM (Big ideas math online) i need help plz

crimeas [40]

Answer: h=80 is the final answer. *The answer must be have a positive sign.*

Explanation:

Switch sides of an equation.

3/4h=60

multiply by 4 both sides of an equation.

4*3/4h=60*4

simplify.

3h=240

divide by 3 both sides of an equation.

3h/3=240/3

simplify.

240/3=80

80*3=240

3*80=240

240/80=3

h=80

Hope this helps!

Thanks!

Have a great day!

5 0

3 years ago

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The product of a number and 9, increased by 4, is 58. Find the number

ella [17]

Answer:

first do 58-4 which is 54 then do 54/ 9 which is 6 so the number is 6

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3 years ago

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Suppose that y varies inversely with x. Use the information to find k, and then choose the equation of variation. x = 1.2 when y

baherus [9]

Y = k/x where k is the constant of variation
y = 3 , x = 1.2 gives
3 = k/1.2
therefore
k = 3*1.2 = 3.6
and the required equation is

y = 3.6 / x

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3 years ago

Which problem solving strategy would be best to solve this problem?

g100num [7]

Answer:

Step-by-step explanation:

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3 years ago

The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra

Snezhnost [94]

Let $X$ denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by $X_1,\ldots,X_{36}$ , each independently and identically distributed with distribution $X_i\sim\mathcal N(26,7.2)$ .

You want to find

$\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)$

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

$\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)$

Recall that if $X\sim\mathcal N(\mu,\sigma)$ , then the sampling distribution $\overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right)$ with $n$ being the size of the sample.

Transforming to the standard normal distribution, you have

$Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}$

so that in this case,

$Z=6\dfrac{\overline X-26}{7.2}$

and the probability is equivalent to

$\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)$
$=\mathbb P(Z>1.481)\approx0.0693$

5 0

3 years ago