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Aleksandr [31]
3 years ago
10

Help with 10 please!

Mathematics
1 answer:
Alexus [3.1K]3 years ago
8 0
Gabrielle is x years old.
Felicia is 6 years older.....so Felicia is x + 6
the mom is twice as old as felicia...so the mom is 2(x + 6)
the aunt is x years older then mom....so the aunt (tanya) is 2(x + 6) + x

so Tanya/s age is :
2(x + 6) + x =
2x + 12 + x =
13x + 12 <===
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Please help me solve this​
GalinKa [24]
Complementary angles add up to 90 degrees.

Which means.. the equation is
8x-23+7x-7=90
Combine like terms you should get
15x-30=90
Add 30 on both sides
15x=120
Divide 15 x=8
Plug in 8 to (8x-23)
8(8)=64-23=41
m
8 0
3 years ago
What is the slope of line LM given L(9, -2) and M(3, -5)?
jasenka [17]

Answer:

1/2

Step-by-step explanation:

(y2-y1)/(x2-x1) is the equation for finding slope with two given points

-5-(-2) over 3-9 equals

-3/-6=-1/-2=1/2

7 0
4 years ago
Use the given information to find the p​-value. ​Also, use a 0.05 significance level and state the conclusion about the null hyp
Hatshy [7]

Answer:

We can assume that the statistic is z_{calc}=0.78

p_v = 2* P(z>0.78) = 0.435

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 true proportion of interest is not different from 3/5

Step-by-step explanation:

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is equal to 3/5 or not.:  

Null hypothesis:p=3/5  

Alternative hypothesis:p \neq 3/5  

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  

We can assume that the statistic is z_{calc}=0.78

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>0.78) = 0.435

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 true proportion of interest is not different from 3/5

7 0
3 years ago
The price for certain produce items at Greg's local supermarket are
Vaselesa [24]
72.28 i took the test it was easy bro
8 0
3 years ago
In a survey of 1000 large corporations, 260 said that, given a choice between a job candidate who smokes and an equally qualifie
Usimov [2.4K]

Answer:

The 95% confidence interval for the proportion of corporations preferring a nonsmoking candidate is (0.2328, 0.2872)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 1000, p = \frac{260}{1000} = 0.26

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.26 - 1.96\sqrt{\frac{0.26*0.74}{1000}} = 0.2328

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.26 + 1.96\sqrt{\frac{0.26*0.74}{1000}} = 0.2872

The 95% confidence interval for the proportion of corporations preferring a nonsmoking candidate is (0.2328, 0.2872)

4 0
3 years ago
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