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stich3 [128]
2 years ago
8

Triangle A B C is shown. Angle A C B is a right angle. The length of hypotenuse A B is 12 centimeters, the length of C B is 9.8

centimeters, and the length of A C is 6.9 centimeters.
Which expressions can be used to find m∠BAC? Select three options.

cos−1(StartFraction 6.9 Over 12 EndFraction)
cos−1(StartFraction 9.8 Over 12 EndFraction)
sin−1(StartFraction 6.9 Over 12 EndFraction)
sin−1(StartFraction 9.8 Over 12 EndFraction)
tan−1(StartFraction 6.9 Over 9.8 EndFraction)

Mathematics
2 answers:
Dafna11 [192]2 years ago
3 0

Step-by-step explanation:

yo digo que es eso la verdad no se

Tcecarenko [31]2 years ago
3 0

Answer:

pls mark brainliest

Step-by-step explanation

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A study was conducted to determine if there was a difference in the driving ability of students from West University and East Un
Sholpan [36]

Answer:

There is no significant evidence which shows that there is a difference in the driving ability of students from West University and East University, <em>assuming a significance level 0.1</em>

Step-by-step explanation:

Let p1 be the proportion of West University students who involved in a car accident within the past year

Let p2 be the proportion of East University students who involved in a car accident within the past year

Then

H_{0}:p1=p2

H_{a}:p1≠p2

The formula for the test statistic is given as:

z=\frac{p1-p2}{\sqrt{\frac{p*(1-p)*(n1+n2)}{n1*n2} } }  where

  • p1 is the <em>sample</em> proportion of West University students who involved in a car accident within the past year (0.15)
  • p2 is the <em>sample</em> proportion of East University students who involved in a car accident within the past year (0.12)
  • p is the pool proportion of p1 and p2 (\frac{15+12}{100+100}=0.135)
  • n1 is the sample size of the students from West University (100)
  • n2 is the sample size ofthe students from East University (100)

Then we have z=\frac{0.03}{\sqrt{\frac{0.135*0.865*(100+100)}{100*100} } } ≈ 0.6208

Since this is a two tailed test, corresponding p-value for the test statistic is ≈ 0.5347.

<em>Assuming significance level 0.1</em>, The result is not significant since 0.5347>0.1. Therefore we fail to reject the null hypothesis at 0.1 significance

6 0
2 years ago
An investor believes that investing in domestic and international stocks will give a difference in the mean rate of return. They
arlik [135]

Answer:

So on this case the 90% confidence interval would be given by -4.137 \leq \mu_1 -\mu_2 \leq 2.087  

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X_1 =2.0233 represent the sample mean 1  

\bar X_2 =3.048 represent the sample mean 2  

n1=15 represent the sample 1 size  

n2=15 represent the sample 2 size  

s_1 =4.893387 population sample deviation for sample 1  

s_2 =5.12399 population sample deviation for sample 2  

\mu_1 -\mu_2 parameter of interest at 0.1 of significance so the confidence would be 0.9 or 90%

We want to test:

H0: \mu_1 = \mu_2

H1: \mu_1 \neq \mu_2

And we can do this using the confidence interval for the difference of means.

Solution to the problem  

The confidence interval for the difference of means is given by the following formula:  

(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}} (1)  

The point of estimate for \mu_1 -\mu_2 is just given by:  

\bar X_1 -\bar X_2 =2.0233-3.048=-1.0247

The degrees of freedom are given by:

df = n_1 +n_2 -2 = 15+15-2=28  

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,28)".And we see that t_{\alpha/2}=\pm 1.701  

Now we have everything in order to replace into formula (1):  

-1.0247-1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=-4.137  

-1.0247+1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=2.087  

So on this case the 90% confidence interval would be given by -4.137 \leq \mu_1 -\mu_2 \leq 2.087  

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

4 0
2 years ago
−3 1/3 ÷ 9 pllllllllllllllllllllllllllllllllllllllllllllllllll
tia_tia [17]

-\frac{10}{27}  hope this helps :)

8 0
2 years ago
If you need to cut 5 pieces of glass from a 14 feet length, how long should each piece be
ivanzaharov [21]
2.8 feet because 14 divided by 5 equals 2.8
5 0
1 year ago
What is the smallest positive multiple of $32$?
Artemon [7]

Answer:

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explanation:

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