How do you do this? Explanation please #46

55,37, 52,50, 53,46,53 whats the standard deviation of the data set?

Andreas93 [3]

Answer:

5.7285536159324

7 0

3 years ago

Mrs. Stern washed

bekas [8.4K]

Answer:

Mrs. Stern washed more

$\frac{7}{20}$ laundry still needs to be washed.

Step-by-step explanation:

Total amount of laundry: $\frac{20}{20}$

$\frac{20}{20} -\frac{2}{5}-\frac{1}{4} =$

$\frac{20}{20} - \frac{8}{20} -\frac{5}{20} =$

$\frac{7}{20}$ of the laundry still needs to be done.

Mrs. Stern washed $\frac{8}{20}$ of the laundry.

Her son washed $\frac{5}{20}$ of the laundry.

Mrs. Stern washed more laundry than her son.

8 0

3 years ago

2 + 2 - 1 I really need help

shutvik [7]

Answer:

3

Step-by-step explanation:

Break the problem up into two

Ignore the -1 for now

2+2=4 because if you separate 2 and 2, you get four ones, therefore it's four

Now that you have four, subtract 1

And you got 3

5 0

3 years ago

When a scientist conducted a genetics experiments with peas, one sample of offspring consisted of 943 peas, with 717 of them hav

pshichka [43]

Using the normal approximation to the binomial distribution, it is found that:

a) 0.242 = 24.2% probability of getting 717 or more peas with red flowers.

b) Since Z < 2, 717 peas with red flowers is not significantly high.

c) Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.

For each pea, there are only two possible outcomes. Either they have a red flower, or they do not. The probability of a pea having a red flower is independent of any other pea, which means that the binomial distribution is used to solve this question.

Binomial distribution:

Probability of x successes on n trials, with p probability.

Normal distribution:

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

If Z > 2, the result is considered significantly high.

If $np \geq 10$ and $n(1-p) \geq 10$ , the binomial distribution can be approximated to the normal with:

$\mu = np$

$\sigma = \sqrt{np(1-p)}$

In this problem:

943 peas, thus, $n = 943$
3/4 probability of being red, thus $p = \frac{3}{4} = 0.75$ .

Applying the approximation:

$\mu = np = 943(0.75) = 707.25$

$\sigma = \sqrt{np(1-p)} = \sqrt{943(0.75)(0.25)} = 13.297$

Item a:

Using continuity correction, this probability is $P(X \geq 717 - 0.5) = P(X \geq 716.5)$ , which is 1 subtracted by the p-value of Z when X = 716.5.

Then:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{716.5 - 707.25}{13.297}$

$Z = 0.7$

$Z = 0.7$ has a p-value of 0.758.

1 - 0.758 = 0.242

0.242 = 24.2% probability of getting 717 or more peas with red flowers.

Item b:

Since Z < 2, 717 peas with red flowers is not significantly high.

Item c:

Since 717 peas with red flowers is not a significantly high result, we cannot conclude that the scientist's assumption is wrong.

A similar problem is given at brainly.com/question/25212369

6 0

3 years ago

Can someone please solve this one? You can just give the answer and no explanation, I just need to get this in real quick.

Inessa05 [86]

Answer:

Step-by-step explanation:

1. T

2. F

3. F

4. T

5. F

6. T

3 0

3 years ago