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3 years ago

13

Asphere has a diameter of 10 in. What is the volume of the sphere?

1 answer:

Pavel [41]3 years ago

3 0

Answer: $\frac{500}{3}$ π $in^{3}$

Step-by-step explanation: diameter=10 in

radius= $\frac{diameter}{2}$ = $\frac{10}{2}$ =5 in

volume= $\frac{4}{3}$ $r^{3}$

= $\frac{4}{3} X 5^{3}$ x π = $\frac{500}{3}$ π $in^{3}$

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5 1/6 + (7 1/3)l <60

agasfer [191]

Answer:

$l$

Step-by-step explanation:

Given the inequality

$5\dfrac{1}{6}+7\dfrac{1}{3}l$

First, convert mixed numbers to improper fractions:

$5\dfrac{1}{6}=\dfrac{5\cdot 6+1}{6}=\dfrac{31}{6}\\ \\7\dfrac{1}{3}=\dfrac{7\cdot 3+1}{3}=\dfrac{22}{3}$

Now the inequality looks like

$\dfrac{31}{6}+\dfrac{22}{3}l$

Multiply it by 6:

$\dfrac{31}{6}\cdot 6+\dfrac{22}{3}l\cdot 6$

7 0

3 years ago

If A and B are independent events, P(A) = 0.25, and P(B) = 0.3, what is P(AB)?

Marrrta [24]

Answer:

$P(A) = 0.25, P(B= 0.3$

And if we want to find $P(A \cap B)$ we can use this formula from the definition of independent events :

$P(A \cap B) =P(A) *P(B) = 0.25*0.3= 0.075$

And the best option would be:

$P(A \cap B) =0.075$

Step-by-step explanation:

For this case we have the following events A and B and we also have the probabilities for each one given:

$P(A) = 0.25, P(B= 0.3$

And if we want to find $P(A \cap B)$ we can use this formula from the definition of independent events :

$P(A \cap B) =P(A) *P(B) = 0.25*0.3= 0.075$

And the best option would be:

$P(A \cap B) =0.075$

3 0

3 years ago

2) An $80 text book was on sale for $49.60 at the used book store. By

nadya68 [22]

Answer:

38%

Step-by-step explanation:

Original price = $ 80

Price after reduction = $ 49.60

Reduction = 80 - 49.60 = $30.40

Percentage of decrease = $\frac{reduction price}{original price}*100$

$= \frac{30.40}{80}*100\\\\=\frac{3040}{80}$

= 38%

3 0

2 years ago

Read 2 more answers

Angle CDE has a slightly greater measure than the measure

svlad2 [7]

Answer:

k

Step-by-step explanation:

6 0

2 years ago

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Working at home: According to the U.S Census Bureau, 41% of men who worked at home were college graduates. In a sample of 506 wo

sergejj [24]

Solution :

a). The point estimate of proportion of college graduates among women who work at home,

$\hat p =\frac{166}{506}$

= 0.3281

99.5% confidence interval

$=\left( \hat p \pm Z_{0.005/2} \sqrt{\frac{\hat p (1- \hat p)}{n}} \right)$

$=\left( 0.3281 \pm 2.81 \sqrt{\frac{0.3281 \times (1- 0.3281)}{506}} \right)$

$=(0.3281 \pm 0.0586)$

$=(0.2695, 0.3867)$

6 0

3 years ago