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Kacy is on her freshman volleyball team.she makes 31%of her foul shots,leading her team to 14 wins in 19 games of the season. If

LUCKY_DIMON [66]

Answer:

15.2

Step-by-step explanation:

She makes 31% of her foul shots

she took 49 shots

the number of foul shots is 31% of 49 shots

=31% * 49

=0.31 * 49

= 15.19

5 0

3 years ago

Suppose j varies jointly with g and v, and j=2 when g=4 and v=3. Find j when g=8 and v=9.

vladimir2022 [97]

Answer:

$j=12$

Step-by-step explanation:

we know that

In a join variation, If j varies jointly with respect to g and v, the equation will be of the form $j = kgv$

where k is a constant

step 1

Find the value of k

we have

j=2,g=4,v=3

substitute and solve for k

$2 = k(4)(3)$

$2 = 12k$

$k=\frac{1}{6}$

The equation is equal to

$j=\frac{1}{6}gv$

step 2

Find the value of j when g=8,v=9

substitute the values in the equation and solve for j

$j=\frac{1}{6}(8)(9)$

$j=12$

3 0

3 years ago

First rewrite 10/21 and 4/9 so that they have a common denominator

Tema [17]

Answer:

10/21=30/63

4/9=28/63

Step-by-step explanation:

Find the lcm (least common multiple) of 21 an 9, which is 63. Then divide 21 and 9 separately by 63. The multiply that number by the numerator.

6 0

1 year ago

The heights of 40 randomly chosen men are measured and found to follow a normal distribution. An average height of 175 cm is obt

AVprozaik [17]

Answer:

95% two-sided confidence interval for the true mean heights of men is [168.8 cm , 181.2 cm].

Step-by-step explanation:

We are given that the heights of 40 randomly chosen men are measured and found to follow a normal distribution.

An average height of 175 cm is obtained. The standard deviation of men's heights is 20 cm.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

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

where, $\bar X$ = sample average height = 175 cm

$\sigma$ = population standard deviation = 20 cm

n = sample of men = 40

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

So, 95% confidence interval for the true mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ]

= [ $175-1.96 \times }{\frac{20}{\sqrt{40} } }$ , $175+1.96 \times }{\frac{20}{\sqrt{40} } }$ ]

= [168.8 cm , 181.2 cm]

Therefore, 95% confidence interval for the true mean height of men is [168.8 cm , 181.2 cm].

The interpretation of the above interval is that we are 95% confident that the true mean height of men will be between 168.8 cm and 181.2 cm.

3 0

3 years ago

What is the volume of a cone with a radius of 6cm and a height of 12cm?

kompoz [17]

Answer:

The answer is 452.16 cm³.

Formula to calculate the volume of cone = V = πr²x h/3

Now r = 6 cm and h = 12cm, and value of π = 3.14

By putting the values,

V = (3.14) (6)² (12/3) = (3.14)(36)(4)

5 0

3 years ago

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