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Tpy6a [65]
3 years ago
10

Can I please have help

Mathematics
1 answer:
earnstyle [38]3 years ago
8 0
Hi,

Formula for ball volume (considered a sphera) is:

V=\dfrac{4\pi\cdot r^3}3=\dfrac{4\pi\cdot 15^3}3=\dfrac{42,390}3=14,130\ in^3.

Green eyes.
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Can anyone help me with my math homework plz
Svetach [21]

Answer:

1. 136

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4. 110

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8 0
3 years ago
Find the midpoint M of the line segment joining the points A = (-5, 7) and B = (-1, -5).
11111nata11111 [884]

Answer:

M = (-3,1)

Step-by-step explanation:

We need to find the midpoint M of the line segment joining the points

A = (-5,7) and B = (-1, -5)

The mid point of the line segment joining the points (x₁,y₁) and (x₂,y₂) is given by :

M=(\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2})\\\\M=(\dfrac{-5+(-1)}{2},\dfrac{7+(-5)}{2})\\\\M=(-3,1)

Hence, the required point is (-3,1).

6 0
3 years ago
supervisor records the repair cost for 25 randomly selected dryers. A sample mean of $93.36 and standard deviation of $19.95 are
Simora [160]
<h2><u>Answer with explanation:</u></h2>

The confidence interval for population mean (when population standard deviation is unknown) is given by :-

\overline{x}-t^*\dfrac{s}{\sqrt{n}}< \mu

, where n= sample size

\overline{x} = Sample mean

s= sample size

t* = Critical value.

Given : n= 25

Degree of freedom : df=n-1=24

\overline{x}= \$93.36

s=\ $19.95

Significance level for 98% confidence interval : \alpha=1-0.98=0.02

Using t-distribution table ,

Two-tailed critical value for 98% confidence interval :

t^*=t_{\alpha/2,\ df}=t_{0.01,\ 24}=2.4922

⇒ The critical value that should be used in constructing the confidence interval = 2.4922

Then, the 95% confidence interval would be :-

93.36-(2.4922)\dfrac{19.95}{\sqrt{25}}< \mu

=93.36-9.943878< \mu

=93.36-9.943878< \mu

=83.416122< \mu

Hence, the 98% confidence interval for the mean repair cost for the dryers. = 83.4161

4 0
3 years ago
Please help me I don’t understand this
Tomtit [17]

Answer:64

Step-by-step explanation:

67

4 0
2 years ago
A large university is interested in learning about the average time it takes students to drive to campus. The university sampled
Dmitrij [34]

Answer: The value of test statistic is 2.442.

Step-by-step explanation:

Since we have given that

n = 238

Sample mean = 23.2 minutes

Standard deviation = 20.26 minutes

Hypothesis :

H_0:\mu=20\\H_a:\mu\neq 20

So, the test statistic value would be

z=\dfrac{\bar{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\\\\z=\dfrac{23.2-20}{\dfrac{20.26}{\sqrt{236}}}\\\\z=2.442

Hence, the value of test statistic is 2.442.

4 0
3 years ago
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