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3 years ago

15

Find the size of the angle a.Diagram not drawn accurately

1 answer:

pickupchik [31]3 years ago

6 0

Answer:

45°

Step-by-step explanation:

Since,

3a + a + 2a + 2a = 360°

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What is the equation of a line that passes through the point (6, 1) and is perpendicular to the line whose equation is y=−2x−6y=

Mandarinka [93]

Lets get the negative reciprocal of the slope so we get
$m = \frac{1}{2}$
Now the equation that is perpendicular is
$y = \frac{1}{2} x + b$
Lets solve for b by plugging in the point (x,y)=(6,1)
$1 = \frac{1}{2} (6) + b \\ 1 = 3 + b \\ 1 - 3 = b \\ b = - 2$
Now plug that into the equation. So the equation of the line that is perpendicular and
passes through the point (6, 1) is
$y = \frac{1}{2} x - 2$

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3 years ago

In ΔNOP, the measure of ∠P=90°, NO = 96 feet, and OP = 36 feet. Find the measure of ∠O to the nearest tenth of a degree.

sladkih [1.3K]

Answer:

68

Step-by-step explanation:

4 0

3 years ago

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol

uysha [10]

Answer:

$(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743$

$(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926$

And we are 9% confidence that the true mean for the difference of the population means is given by:

$0.0743 \leq \mu_1 -\mu_2 \leq 1.926$

Step-by-step explanation:

For this problem we have the following data given:

$\bar X_1 = 9$ represent the sample mean for one of the departments

$\bar X_2 = 8$ represent the sample mean for the other department

$n_1 = 25$ represent the sample size for the first group

$n_2 = 20$ represent the sample size for the second group

$s_1 = 2$ represent the deviation for the first group

$s_2 =1$ represent the deviation for the second group

Confidence interval

The confidence interval for the difference in the true means is given by:

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

The confidence given is 95% or 9.5, then the significance level is $\alpha=0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by:

$df=n_1 +n_2 -2= 20+25-2= 43$

And the critical value for this case is $t_{\alpha/2}=2.02$

And replacing we got:

$(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743$

$(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926$

And we are 9% confidence that the true mean for the difference of the population means is given by:

$0.0743 \leq \mu_1 -\mu_2 \leq 1.926$

4 0

3 years ago

Find the missing side of the triangle. Side LaTeX: cc is the hypotenuse and sides LaTeX: aa and LaTeX: bb are legs. Leave your a

mylen [45]

Answer:

$c = 2\sqrt{5$

Step-by-step explanation:

Given

$a = 2$

$b = 4$

c = hypotenuse

Required

Find c

Using Pythagoras theorem, we have:

$c^2 = a^2 + b^2$

Substitute values for a and b

$c^2 = 2^2 + 4^2$

$c^2 = 4 + 16$

$c^2 = 20$

Take square roots of both sides:

$c = \sqrt{20$

Express 20 as 4 * 5

$c = \sqrt{4 * 5$

Split surd

$c = \sqrt{4} * \sqrt{5$

Take square root of 4

$c = 2 * \sqrt{5$

$c = 2\sqrt{5$

6 0

3 years ago

How many minutes is in 1/4 of a hour

Radda [10]

The answer is 15 minutes

7 0

3 years ago

Read 2 more answers