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2 years ago

Is this aa theorem or sas theorem

Mathematics

1 answer:

sergey [27]2 years ago

8 0

I believe this is SAS because the angles are not the same so it cannot be angle angle it has to be SAS

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Please help me with question #13

lesantik [10]

When dividing these, the powers are subtracted form each other.
We could split up the fraction into r's, s's, and t's.
r^-3 / r^2 = r^-5 [think of it as -3 -2]
s^5 / s = s^4
t^2 / t^-2 = t^4 [2 - - 2 = 4]

A negative power places it on the bottom of the fraction and the minus sign is removed from the power.
So the answer is A: (s^4)(t^4)/r^5

4 0

2 years ago

In 2015 as part of the General Social Survey, 1289 randomly selected American adults responded to this question:

Radda [10]

Answer:

B (0.312, 0.364)

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 interval $1-\alpha$ , 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:

1289 randomly selected American adults responded to this question. This means that $n = 1289$ .

Of the respondents, 436 replied that America is doing about the right amount. This means that $\pi = \frac{436}{1289} = 0.3382$ .

Determine a 95% confidence interval for the proportion of all the registered voters who will vote for the Republican Party. 

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:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3382 - 1.96\sqrt{\frac{0.3382*0.6618}{1289}} = 0.312$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.3382 + 1.96\sqrt{\frac{0.3382*0.6618}{1289}} = 0.364$

The 95% confidence interval is:

B (0.312, 0.364)

5 0

2 years ago

Simplify the following expression.<br> -14x^2+9x-4x^2+5

Naya [18.7K]

Answer:

i think the answer is -18x^(2)+9x+5

-18x^(2)+9x+5

Subtract 4x^2 from −14x^2

3 0

3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

3 years ago

If log2 5 = k, determine an expression for log32 5 in terms of k.

lukranit [14]

Answer:

$log_3_2(5)=\frac{1}{5} k$