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Natali5045456 [20]
3 years ago
9

ITS DUE TODAY! HELP ME PLEASE

Mathematics
1 answer:
Brums [2.3K]3 years ago
8 0

Mean - 4.19

median - 4

modes - 0,1,1,2,2,3,3,3,3,3,4,4,5,5,5,6,7,7,7,8,9

range - 9

interquartile range - 4

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A person runs 1/8 mile in 1/64 hour.<br> The person's speed is (BLANK) miles per hour.
snow_tiger [21]

Answer: The person's speed is 8 miles per hour.

Step-by-step explanation:

To solve we want to convert the \frac{1}{8} and \frac{1}{64} proportionally for the number of miles per hour. We know that multiplying a fraction by its reciprocal will equal 1. So let's do that for the value of the hour.

\frac{1}{64}×\frac{64}{1} =\frac{64}{64} =1

Since we multiplied that to the number of hours we must also do that for the number of miles.

\frac{1}{8}×\frac{64}{1} =\frac{64}{8} =8

7 0
2 years ago
Let us suppose we have data on the absorbency of paper towels that were produced by two different manufacturing processes. From
maksim [4K]

Answer:

The 95% CI for the difference of means is:

-155.45 \leq \mu_1-\mu_2 \leq -44.55

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>"Find a 95% confidence interval on the difference of the towels mean absorbency produced by the two processes. Assumed that the standard deviations are estimated from the data. Round to two decimals places."</em>

Process 1:

- Sample size: 10

- Mean: 200

- S.D.: 15

Process 2:

- Sample size:  4

- Mean: 300

- S.D.: 50

The difference of the sample means is:

M_d=M_1-M_2=200-300=-100

The standard deviation can be estimated as:

\sigma_d=\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\\\\sigma_d=\sqrt{\frac{15^2}{10}+\frac{50^2}{4}} =\sqrt{22.5+625}=\sqrt{647.5}=25.45

The degrees of freedom are:

df=n_1+n_2-2=10+4-2=12

The t-value for a 95% confidence interval and 12 degrees of freedom is t=±2.179.

Then, the confidence interval can be written as:

M_d-t\cdot \sigma_d\leq \mu_1-\mu_2 \leq M_d+t\cdot \sigma_d\\\\-100-2.179*25.45\leq \mu_1-\mu_2 \leq -100+2.179*25.45\\\\-100-55.45 \leq \mu_1-\mu_2 \leq -100+55.45\\\\ -155.45 \leq \mu_1-\mu_2 \leq -44.55

8 0
3 years ago
Find the equivalent to x^7
MrMuchimi
1/x^-7

It would look more like       1
                                        ------
                                             -7 (as an exponent)
                                          x
3 0
3 years ago
Suppose that x has a binomial distribution with n = 201 and p = 0.45. (Round np and n(1-p) answers to 2 decimal places. Round yo
ycow [4]

Answer:

a) It can be used because np and n(1-p) are both greater than 5.

Step-by-step explanation:

Binomial distribution and approximation to the normal:

The binomial distribution has two parameters:

n, which is the number of trials.

p, which is the probability of a success on a single trial.

If np and n(1-p) are both greater than 5, the normal approximation to the binomial can appropriately be used.

In this question:

n = 201, p = 0.45

So, lets verify the conditions:

np = 201*0.45 = 90.45 > 5

n(1-p) = 201*(1-0.45) = 201*0.55 = 110.55 > 5

Since both np and n(1-p) are greater than 5, the approximation can be used.

3 0
3 years ago
What is the value of 4.5 x 10^5 written in standard form?
soldier1979 [14.2K]

Answer:

4500000

Step-by-step explanation: Let me know if this helped

3 0
3 years ago
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