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Nikolay [14]
3 years ago
15

Hector enters a competition to guess how many stuffed animals are in a box. Hector’s guess is 35 stuffed animals. The actual num

ber of stuffed animals is 28. What is the percent error of Hector’s guess?
Mathematics
1 answer:
qwelly [4]3 years ago
5 0

Answer:20%

Step-by-step explanation:Based on the given conditions, formulate:( 35 - 28 ) divided by 35

Calculate the sum or difference: 7/ 35

Cross out the common factor: 1/5

Multiply a number to both the numerator and the denominator: 1/5 x 20/20

Write as a single fraction: 20/5x20

Calculate the product or quotient:20/100

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The students at a High School earned money for an international animal rescue foundation. 82 seniors earned an average $26.75 pe
Anon25 [30]

Answer: A. $16.13

Step-by-step explanation:

Total students = 82+74+96+99 =351

Sum of earnings of 82 seniors  = $26.75  x 82= $2193.5

Sum of earnings of 74 juniors  = $12.25 x 74 = $906.5

Sum of earnings of 96 sophomores  = $15.50  x 96 = $1488

Sum of earnings of 99 freshmen  =  $10.85 x 99 = $1074.15

Total earnings =  $2193.5  + $906.5+  $1488 +$1074.15

= $5662.15

(Total earnings) ÷ (Total students )

= $5662.15÷ 351

= $16.13

4 0
3 years ago
What is the formula for surface area
zubka84 [21]
The area is perimeter times height plus area of the bases
6 0
3 years ago
Read 2 more answers
I need help please quiz
Anon25 [30]

Answer:

f(x - 4) = x^2 - 6x+ 8

Step-by-step explanation:

Given

f(x) = x^2 + 2x

Required

Find f(x - 4)

Substitute x - 4 for x

f(x - 4) = (x- 4)^2 + 2(x - 4)

Expand

f(x - 4) = x^2 - 8x + 16 + 2x - 8

Collect like terms

f(x - 4) = x^2 - 8x + 2x+ 16  - 8

f(x - 4) = x^2 - 6x+ 8

4 0
3 years ago
Estimate the sum of 23 and 71
Natali [406]
Estimation:

23 --> rounded to 20
71 --> rounded to 70
70 + 20 = 90. I rounded both numbers down so the exact sum must be only a little greater than my estimate.
8 0
3 years ago
Read 2 more answers
According to an article in Newsweek, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100:114 (46.7%
mojhsa [17]

Answer:

z=\frac{0.42 -0.467}{\sqrt{\frac{0.467(1-0.467)}{150}}}=-1.154  

p_v =2*P(z  

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of girls born is not significantly different from 0.467

Step-by-step explanation:

Data given and notation

n=150 represent the random sample taken

X=63 represent the number of girls born

\hat p=\frac{63}{150}=0.42 estimated proportion of girls born

p_o=0.467 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion if girls is 0.467.:  

Null hypothesis:p=0.467  

Alternative hypothesis:p \neq 0.467  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

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

z=\frac{0.42 -0.467}{\sqrt{\frac{0.467(1-0.467)}{150}}}=-1.154  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a bilateral test the p value would be:  

p_v =2*P(z  

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of girls born is not significantly different from 0.467

3 0
3 years ago
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