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Ymorist [56]
3 years ago
10

What is the radius of a sphere with a volume of 7961 cm", to the nearest tenth of a centimeter?

Mathematics
1 answer:
statuscvo [17]3 years ago
6 0

Answer:

12.4 cm

Step-by-step explanation:

Use the sphere volume formula, V = \frac{4}{3}\pir³

Plug in the volume and solve for r, the radius:

V = \frac{4}{3}\pir³

7961 = \frac{4}{3}\pir³

1900.55 = r³

12.39 = r

So, the radius of the sphere is approximately 12.4 cm

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Emmanuel added two integers Which condition will always give Emmanuel a negative solution when he adds two integers? Both intege
alexgriva [62]

Both integers have negative values

Explanation

Let

x and y represents the number

so, the options are

\begin{gathered} +\text{   +     +} \\ +\text{  +      -} \\ -\text{     +     +} \\ -\text{   +-} \end{gathered}

a) Both integers have negative values Both integers

-x-y=-(x+y)\rightarrow you\text{ will always get a negative number}

b) Both integers have positive values

x+y=+\text{  +  += you will always get a positive number}

c)One integer has a positive value, and one integer has a negative value

\begin{gathered} -x+y\text{  or x-y} \\ the\text{ sign of the result depends on the greates absolute value, it means the answer will have the same of the bigger number( absolute value)} \end{gathered}

d)The values of the two integers are opposites​

as the point C , the sign depends on the bigger number

for example

\begin{gathered} 8-5=3\text{ positive because 8 is the bigger} \\ 5-8=-3\text{ negative because (-8) is has the bigger absolute value} \end{gathered}

so, the answer is Both integers have negative values

I hope this helps you

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−5(x+2)=−5x+n<br> What is n?
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I’m pretty sure that n is equal to -10
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n a survey of a group of​ men, the heights in the​ 20-29 age group were normally​ distributed, with a mean of inches and a stand
kotykmax [81]

Answer:

(a) The probability that a study participant has a height that is less than 67 inches is 0.4013.

(b) The probability that a study participant has a height that is between 67 and 71 inches is 0.5586.

(c) The probability that a study participant has a height that is more than 71 inches is 0.0401.

(d) The event in part (c) is an unusual event.

Step-by-step explanation:

<u>The complete question is:</u> In a survey of a group of​ men, the heights in the​ 20-29 age group were normally​ distributed, with a mean of 67.5 inches and a standard deviation of 2.0 inches. A study participant is randomly selected. Complete parts​ (a) through​ (d) below. ​(a) Find the probability that a study participant has a height that is less than 67 inches. The probability that the study participant selected at random is less than inches tall is nothing. ​(Round to four decimal places as​ needed.) ​(b) Find the probability that a study participant has a height that is between 67 and 71 inches. The probability that the study participant selected at random is between and inches tall is nothing. ​(Round to four decimal places as​ needed.) ​(c) Find the probability that a study participant has a height that is more than 71 inches. The probability that the study participant selected at random is more than inches tall is nothing. ​(Round to four decimal places as​ needed.) ​(d) Identify any unusual events. Explain your reasoning. Choose the correct answer below.

We are given that the heights in the​ 20-29 age group were normally​ distributed, with a mean of 67.5 inches and a standard deviation of 2.0 inches.

Let X = <u><em>the heights of men in the​ 20-29 age group</em></u>

The z-score probability distribution for the normal distribution is given by;

                          Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean height = 67.5 inches

            \sigma = standard deviation = 2 inches

So, X ~ Normal(\mu=67.5, \sigma^{2}=2^{2})

(a) The probability that a study participant has a height that is less than 67 inches is given by = P(X < 67 inches)

 

      P(X < 67 inches) = P( \frac{X-\mu}{\sigma} < \frac{67-67.5}{2} ) = P(Z < -0.25) = 1 - P(Z \leq 0.25)

                                                                 = 1 - 0.5987 = <u>0.4013</u>

The above probability is calculated by looking at the value of x = 0.25 in the z table which has an area of 0.5987.

(b) The probability that a study participant has a height that is between 67 and 71 inches is given by = P(67 inches < X < 71 inches)

    P(67 inches < X < 71 inches) = P(X < 71 inches) - P(X \leq 67 inches)

    P(X < 71 inches) = P( \frac{X-\mu}{\sigma} < \frac{71-67.5}{2} ) = P(Z < 1.75) = 0.9599

    P(X \leq 67 inches) = P( \frac{X-\mu}{\sigma} \leq \frac{67-67.5}{2} ) = P(Z \leq -0.25) = 1 - P(Z < 0.25)

                                                                = 1 - 0.5987 = 0.4013

The above probability is calculated by looking at the value of x = 1.75 and x = 0.25 in the z table which has an area of 0.9599 and 0.5987 respectively.

Therefore, P(67 inches < X < 71 inches) = 0.9599 - 0.4013 = <u>0.5586</u>.

(c) The probability that a study participant has a height that is more than 71 inches is given by = P(X > 71 inches)

 

      P(X > 71 inches) = P( \frac{X-\mu}{\sigma} > \frac{71-67.5}{2} ) = P(Z > 1.75) = 1 - P(Z \leq 1.75)

                                                                 = 1 - 0.9599 = <u>0.0401</u>

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.9599.

(d) The event in part (c) is an unusual event because the probability that a study participant has a height that is more than 71 inches is less than 0.05.

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Step-by-step explanation:

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