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stiv31 [10]
2 years ago
9

I’m about to cry from frustration. Help please before I die

Mathematics
2 answers:
kvv77 [185]2 years ago
7 0

Answer:

\frac{1}{3}x+16=37

63 Jelly Beans

Step-by-step explanation:

The unknown is the number of jelly beans originally in the bag or x

First he had x jelly beans

The he ate one-third of them

\frac{1}{3}x

He then ate 16 more jelly beans

\frac{1}{3}x+16

This was equal to 37 jelly beans

\frac{1}{3}x+16=37

This is the equation

Now solve for x

Subtract 16 from both sides

\frac{1}{3}x=21

Multiply both sides by 3

x=63

63 Jelly Beans

Strike441 [17]2 years ago
7 0
First you write the equation x+16=37. After solving you will get x=21. Then you multiply x by 3 to get the whole bag. And you will get 63
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One ballispicked at random from a box containing 3redballs and 1green balls. If X is the number of redballs picked, find the pro
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Answer:

P(X = 0) = 0.25

P(X = 1) = 0.75

Step-by-step explanation:

A probability is the number of desired outcomes divided by the number of total outcomes.

The probability distribution of X.

3 red balls out of 4, one picked, so we can have either 0 red balls or 1 red ball. The probability disitribution is the probability of each outcome.

P(X = 0) = \frac{1}{4} = 0.25

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What is the effect on the area of a rectangle if the dimensions are doubled?
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Suppose 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea. Use a 0.05 significance
lana66690 [7]

Answer:

Null Hypothesis, H_0 : p = 0.20  

Alternate Hypothesis, H_a : p > 0.20  

Step-by-step explanation:

We are given that 241 subjects are treated with a drug that is used to treat pain and 54 of them developed nausea.

We have to use a 0.05 significance level to test the claim that more than 20​% of users develop nausea.

<em>Let p = population proportion of users who develop nausea</em>

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<u>Alternate Hypothesis</u>, H_a : p > 0.20  

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And <u><em>alternate hypothesis</em></u> states that more than 20​% of users develop nausea.

The test statistics that would be used here is <u>One-sample z proportion</u> test statistics.

                     T.S. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }   ~ N(0,1)

where,  \hat p = proportion of users who develop nausea in a sample of 241 subjects =  \frac{54}{241}  

             n = sample of subjects = 241

So, the above hypothesis would be appropriate to conduct the test.

6 0
3 years ago
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