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SpyIntel [72]
3 years ago
5

Sam tested every 50th candy bar from the assembly line to make sure there were enough peanuts in each bar. He found 15% did not

have enough peanuts
How many candy bars do not have enough peanuts if the factory produces 20,000 candy bars?
A 300 candy bars
OB. 15,000 candy bars
o C. 3,000 candy bars
O D. 500 candy bars
Mathematics
2 answers:
Vlad [161]3 years ago
8 0
I think the answer is B
grandymaker [24]3 years ago
5 0

Answer:

C 3,000

Step-by-step explanation:

the step is you would take 15% of 20,000 candy bars

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A recent health report revealed that a woman with insurance spends an average of 2.3 days in the hospital following a routine ch
cupoosta [38]

Answer:

t=\frac{2.3-1.9}{\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}}=2.385  

p_v =2*P(t_{30}>2.385)=0.0236

Comparing the p value with the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis

The confidence interval would be given by:

(2.3 -1.9) - 2.75*\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=-0.0611

(2.3 -1.9) + 2.75* \sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=0.861

Step-by-step explanation:

Data given and notation

\bar X_{insurance}=2.3 represent the mean for insurance

\bar X_{No. ins}=1.9 represent the mean withour insurance

s_{Insurance}=0.6 represent the sample standard deviation for the insurance case

s_{No. Ins}=0.3 represent the sample standard deviation for the No insurance case

n_{Insurance}=16 sample size for insurance

n_{No. Iss}=16 sample size for no insurance

t would represent the statistic (variable of interest)

\alpha=0.01 significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the means for the two groups are different, the system of hypothesis would be:

Null hypothesis:\mu_{Insu}=\mu_{No. Ins}

Alternative hypothesis:\mu_{Ins} \neq \mu_{No. Ins}

Since we know the population deviations for each group, for this case is better apply a z test to compare means, and the statistic is given by:

t=\frac{\bar X_{Ins}-\bar X_{No.Ins}}{\sqrt{\frac{s^2_{Ins}}{n_{Ins}}+\frac{s^2_{No. Ins}}{n_{No. Ins}}}} (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Calculate the value of the test statistic for this hypothesis testing.

Since we have all the values we can replace in formula (1) like this:

t=\frac{2.3-1.9}{\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}}=2.385  

What is the p-value for this hypothesis test?

The degrees of freedom are given by:

df = n_{ins} +n_{No Ins}-2=16+16-2 =30

Since is a bilateral test the p value would be:

p_v =2*P(t_{30}>2.385)=0.0236

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given \alpha=0.01 we see that p_v>\alpha so we can conclude that we FAIL to reject the null hypothesis

The confidence interval would be given by:

(2.3 -1.9) - 2.75*\sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=-0.0611

(2.3 -1.9) + 2.75* \sqrt{\frac{0.6^2}{16}+\frac{0.3^2}{16}}}=0.861

6 0
3 years ago
Find the product -3,-4.-2.5
Assoli18 [71]

Answer:

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Step-by-step explanation:

-3(-4)(-2.5)

Apply rule -(-a) = a

=-3*4*2.5

Multiply the numbers: 3*4*2.5=30

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3 years ago
in a weakly connected graph. it is not possible to reach all other vertices from any one vertext (true) (false)
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True it is true because you can reach the vertext at all times try it out my teacher told me right now check it our


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ELEN [110]

Answer:

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Step-by-step explanation:

by dividing the smallest number from the bigger number to get the last house

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(6y^2-7y-3)-(2y^2-9y-7) simplify
madreJ [45]
6y^2-7y-3-2y^2+9y+7 (pretend that you're multiplying -1 to 2y^2-9y-7 which is why 9y and 7 are positive now).
Now you're able to simplify your problem, so...

6y^2-2y^2= 4y^2
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-3+7=4

4y^2+2y+4

Might want to check my math but that's how you do it.
5 0
3 years ago
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