This is a 9th grade algebra question....

A simple random sample of 44 men from a normally distributed population results in a standard deviation of 10.7 beats per minute

sladkih [1.3K]

Answer:

Step-by-step explanation:

The Hypothesis :

H0 : σ = 10

H1 : σ ≠ 10

The Chisquare statistic, χ² ;

χ² = [(n-1) * s²] ÷ σ²

Sample size, n = 44 ; sample Standard deviation, s = 10.7 ; population standard deviation, σ = 10

χ² = [(44 - 1) * 10.7²] ÷ 10²

χ² = [43 * 114.49] ÷ 100

χ² = [4923.07] ÷ 100

χ² = 49.2307

Using the Pvalue from Chisquare calculator :

degree of freedom, df = n - 1 = 44 - 1 = 43

Pvalue = 0.238

7 0

3 years ago

Five members of the soccer team ordered orange juice. Ten members ordered apple juice. Juice was packaged in 250 mL cartons. Exp

vredina [299]

Your answer would be B.

5 0

4 years ago

PLEASE HELP THIS IS MY LAST QUESTION OF HIGHSCHOOL ALGEBRA!!!!

ludmilkaskok [199]

The second choice 81.67

4 0

3 years ago

Read 2 more answers

Please answer a question the images.

Lyrx [107]

The distribution of the animals in the coral reef is an illustration of probability

<h3>The most likely animal</h3>

The distribution of the animals is given as:

Blue fish = 3

Yellow fish = 7

Sea turtle = 1

Eels = 3

The animal with the highest frequency is the most likely animal.

Hence, the Yellow fish is the most likely animal.

<h3>Which is more likely?</h3>

From the question, we have:

Blue fish = 3

Eels = 3

Both animals have the same frequency.

This means that Riley has equal chance of seeing a blue fish and an eel.

<h3>The probability of seeing an animal with fins</h3>

The animal with fins are:

Blue fish = 3

Yellow fish = 7

So, the probability is:

$p = \frac{3 + 7}{3 + 7 + 1 + 3}$

$p = \frac{10}{14}$

Simplify

$p = \frac 57$

Hence, the probability of seeing an animal with fins is 5/7

<h3>The least likely animal</h3>

The animal with the least frequency is:

Sea turtle = 1

So, the likelihood is:

$p = \frac 1{14}$

Hence, the sea turtle has the least likelihood of 1/14

<h3>Animals with equal likelihood</h3>

The Blue fish and the Eels have the same frequency of 3

This means that animals have equal likelihood

<h3>The probability of seeing each animal</h3>

We have:

Blue fish = 3

Yellow fish = 7

Sea turtle = 1

Eels = 3

Express as probability

Blue fish = 3/14

Yellow fish = 7/14

Sea turtle = 1/14

Eels = 3/14

Express the fractions as decimals

Blue fish = 0.214

Yellow fish = 0.500

Sea turtle = 0.071

Eels = 0.214

So, their likelihoods are:

Blue fish = Equally likely

Yellow fish = Certainly likely

Sea turtle = Unlikely

Eels = Equally likely