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olchik [2.2K]
3 years ago
14

Use exponents to write the numbers 81 and 64 in as many different ways as you can. Then write 64/81 using exponents in as many d

ifferent ways as you can.
Mathematics
1 answer:
Vlad1618 [11]3 years ago
3 0

Answer:

c

Step-by-step explanation:

edge 2021

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Please help I’m in a hurry!!! Thank you
d1i1m1o1n [39]

Answer:

A

Step-by-step explanation:

itʻll be equal to 180

the box means 90 degrees

180-90=90

90-29= 61

2*30+1=61

i hope this helps!

6 0
3 years ago
Read 2 more answers
Suppose that the walking step lengths of adult males are normally distributed with a mean of 2.5 feet and a standard deviation o
Hitman42 [59]

Answer:

0.0668 = 6.68% probability that an individual man’s step length is less than 1.9 feet.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

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

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with a mean of 2.5 feet and a standard deviation of 0.4 feet.

This means that \mu = 2.5, \sigma = 0.4

Find the probability that an individual man’s step length is less than 1.9 feet.

This is the p-value of Z when X = 1.9. So

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

Z = \frac{1.9 - 2.5}{0.4}

Z = -1.5

Z = -1.5 has a p-value of 0.0668

0.0668 = 6.68% probability that an individual man’s step length is less than 1.9 feet.

3 0
3 years ago
Whats the answer to this problem????<br><br><br>the product of 6 and a number p is 24
Ad libitum [116K]
Well the P should probably be 4
7 0
3 years ago
Rebeckah answered 42 of the 60 questions correctly on a test. Which method should she use to find the percent of questions she a
Mars2501 [29]

Answer:

The last option

Step-by-step explanation:

To find a percentage of something you need

\frac{expected}{total}*100

The expected is correct marks, the total is the total amount - then you multiply by 100 to turn the decimal into a percentage.

6 0
3 years ago
A study of 10 different weight loss programs involved 500 subjects. Each of the 10 programs had 50 subjects in it. The subjects
Yakvenalex [24]

Answer:

a) (iii) ANOVA

b) The ANOVA test is more powerful than the t test when we want to compare group of means.

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

If we assume that we have p=10 groups and on each group from j=1,\dots,p=10 we have n_j individuals on each group we can define the following formulas of variation:  

SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2

SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2

SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2

And we have this property

SST=SS_{between}+SS_{within}

Solution to the problem

Part a

(i) confidence interval

False since the confidence interval work just when we have just on parameter of interest, but for this case we have more than 1.

(ii) t-test

Can be a possibility but is not the best method since every time that we conduct a t-test we have a chance that we commit a Type I error.

(iii) ANOVA

This one is the best method when we want to compare more than 1 group of means.

(iv) Chi square

False for this case we don't want to analyze independence or goodness of fit, so this one is not the correct test.

Part b

The ANOVA test is more powerful than the t test when we want to compare  group of means.

8 0
4 years ago
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