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2 years ago

IN THE FIGURE 5.24 TO THE RIGHT ABCD IS A RHOMBUS.SHOW THAT AC IS THE BISECTOR OF <BAD.

Mathematics

1 answer:

Inessa05 [86]2 years ago

4 0

In any rhombus, the four sides are the same length.

Because of this, we know that

AD = AB
CD = BC

Furthermore, AC = AC due to the reflective property

Those three facts then allow us to prove triangle ACD is congruent to triangle ACB through the SSS congruence property. Then from CPCTC, we then know that angle DAC = angle BAC, which is enough to show that AC bisects the angle shown. The term "bisect" means "cut in half".

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Evaluate the following, if a=7, b=3, c=2

Zinaida [17]

Answer:

Step-by-step explanation:

a^2 - 5bc - 1

a = 7

b = 3

c = 2

7^2 - 5*3*2 - 1

49 - 30 - 1

4 0

2 years ago

Previously, an organization reported that teenagers spent 24.5 hours per week, on average, on the phone. The organization thinks

Lena [83]

Answer:

We need to develop a one-tail t-student test ( test to the right )

We reject H₀ we find evidence that student spent more than 24,5 hours on the phone

Step-by-step explanation:

Sample size n = 15 n < 30

And we were asked if the mean is higher than, therefore is a one-tail t-student test ( test to the right )

Population mean μ₀ = 24,5

Sample mean μ = 25,7

Sample standard deviation s = 2

Hypothesis Test:

Null Hypothesis H₀ μ = μ₀

Alternative Hypothesis Hₐ μ > μ₀

t (c) = ?

We will define CI = 95 % then α = 5 % α = 0,05 α/2 = 0,025

n = 15 then degree of freedom df = 14

From t-student table we get: t(c) = 2,1448

And t(s)

t(s) = ( μ - μ₀ ) / s/√n

t(s) = (25,7 - 24,5) /2/√15

t(s) = 2,3237

Now we compare t(c) and t(s)

t(c) = 2,1448 t(s) = 2,3237

t(s) > t(c)

Then we are in the rejection region we reject H₀ we have evidence at 95% of CI that students spend more than 24,5 hours per week on the phone

8 0

2 years ago

What is the answer to f/3 < -2

Hitman42 [59]

Answer:

1 i think

Step-by-step explanation:

hope this helps :) have a nice day :) :)

4 0

3 years ago

52= -5x -3 can you solve for x

Roman55 [17]

Answer:

x = -11

Step-by-step explanation:

First, add three to both sides on the equation. Then, your equation becomes 55= -5x. Finally, divide both sides by -5 to get -11 = x. So, x equals -11. Hope this helps!

4 0

3 years ago

Read 2 more answers

The following data were collected from 12 rain gauges in a park. Build a 95% CI for the mean rainfall at the park.

dybincka [34]

Answer:

Critical values: $t_{\alpha/2}=-2.201$ $t_{1-\alpha/2}=2.201$

95% confidence interval would be given by (3.646;4.472)

Step-by-step explanation:

1) 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".

The data is:

4.65 3.89 2.73 4.35 3.80 4.86 4.33 4.37 4.76 4.05 3.05 3.87

2) Compute the sample mean and sample standard deviation.

In order to calculate the mean and the sample deviation we need to have on mind the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$

=AVERAGE(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

On this case the average is $\bar X= 4.059$

=STDEV.S(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

The sample standard deviation obtained was s=0.6503

3) Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =12 <30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by: