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Masja [62]
3 years ago
10

Which test would be used to show these

Mathematics
1 answer:
Alexandra [31]3 years ago
6 0

Answer:

SAS show these triangles are congurent

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Select a composite number to break into factors. Continue factoring until all factors are prime
Sladkaya [172]

Answer:

2*2  * 2*2   * 2*3

Step-by-step explanation:

96 =16 *6

Break these down, since neither 16 nor 6 are prime

    = 4*4 * 2*3

4 in not prime, but 2 and 3 are prime

   = 2*2  * 2*2   * 2*3

All of these are prime

7 0
3 years ago
2x^4–5x^3+x^2+3x+2=?<br><br> X=5
DanielleElmas [232]

Answer:

2x^4–5x^3+x^2+3x+2=

2 \times 5^4–5 \times 5^3+5^2+3 \times 5+2

= 2 × 625 – 5 × 125 + 25 + 3 × 5 + 2

=1250 – 625 + 25 + 15 +2

= 1292 – 625

= 667

3 0
3 years ago
If Jake buys 3 new shirts, how many of them would you expect to be t-shirts, and how many would you expect to be collared?
Sophie [7]

Answer:

2 t-shirts ; 1 collar shirt

Step-by-step explanation:

We need to first obtain the ratio of T - shirts to collar shirts :

T - shirts = 10

Collar shirts = 5

Ratio = T - shirts / Collar shirts = 10 / 5 = 2 /1 = 2:1

Hence, using the ratio obtained ; if Jake buys 3 new shirts :

Number of T-shirts :

(Ratio of t-shirts / total ratio) * new t-shirts

2/3 * 3 = 2 t-shirts

Collar shirts :

1/3 * 3 = 1

2 t-shirts ; 1 collar shirt

4 0
3 years ago
Does going to a private university increase the chance that a student will graduate with student loan debt? A national poll by t
Snowcat [4.5K]

Answer:

Null hypothesis:p \leq 0.69  

Alternative hypothesis:p > 0.69  

Step-by-step explanation:

1) Data given and notation

n=1500 represent the random sample taken

X represent the number of graduates that had student loan debt in 2014

\hat p=0.71 estimated proportion of adults that said that it is morally wrong to not report all income on tax returns

p_o=0.69 is the value that we want to test

\alpha represent the significance level  

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that if there was a significant increase in the proportion of student loan debt for public and nonprofit colleges in 2014 respect to the value of 2013.:  

Null hypothesis:p \leq 0.69  

Alternative hypothesis:p > 0.69  

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.

3) Calculate the statistic  

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

z=\frac{0.71 -0.69}{\sqrt{\frac{0.69(1-0.69)}{1500}}}=1.675  

4) 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 assumed is \alpha=0.05. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

p_v =P(Z>1.675)=0.047  

If we compare the p value obtained and using the significance level assumed \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults that said that it is morally wrong to not report all income on tax returns  is not significantly higher than 0.69.  

5 0
3 years ago
PLEase Hurry!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
mylen [45]

Answer: 56

Step-by-step explanation: Divide 42 by 3 and then multiply by 4

6 0
3 years ago
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