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3 years ago

I WILL GIVE BRAINLIEST IF YOU GET IT RIGHT!!!!

Mathematics

1 answer:

gizmo_the_mogwai [7]3 years ago

3 0

9514 1404 393

Answer:

(a) In triangles APD and BPC; AD = BC

Step-by-step explanation:

We're not concerned with triangle APB, eliminating the last 2 choices.

We're not concerned with sides AP and BP, as they're not sides of the square, eliminating the second choice.

The choice that makes any sense in the proof is ...

In triangles APD and BPC; AD = BC

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A random sample of 30 households was selected as part of a study on electricity usage, and the number of kilowatt-hours (kWh) wa

grin007 [14]

Answer:

The 98% confidence interval for the mean usage in the March quarter of 2006, in kWh, was (333.87, 416.13).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 30 - 1 = 29

98% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 29 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.98}{2} = 0.99$ . So we have T = 2.462

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.462\frac{91.5}{\sqrt{30}} = 41.13$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 375 - 41.13 = 333.87 kWh

The upper end of the interval is the sample mean added to M. So it is 375 + 41.13 = 416.13 kWh

The 98% confidence interval for the mean usage in the March quarter of 2006, in kWh, was (333.87, 416.13).

6 0

3 years ago

Can i please get help with this question?

Feliz [49]

The first letter because 1•5 = 5. 2•5 equals ten etc

4 0

3 years ago

Question 4 plz show ALL STEPS

scoray [572]

Part (a)

Locate x = -1 on the x axis. Draw a vertical line through this x value until you reach the f(x) curve. Then move horizontally until you reach the y axis. You should arrive at y = 4. Check out the diagram below to see what I mean.

Since f(-1) = 4, this means we can then say

g( f(-1) ) = g( 4 ) = 4

To evaluate g(4), we'll follow the same idea as what we did with f(x). However, we'll start at x = 4 and draw a vertical line until we reach the g(x) curve this time.

<h3>Answer: 4</h3>

==========================================================

Part (b)

We use the same idea as part (a)

f(-2) = 5

g( f(-2) ) = g(5) = 6

<h3>Answer: 6</h3>

==========================================================

Part (c)

Same idea as the last two parts. We start on the inside and work toward the outside. Keep in mind that g(x) is now the inner function for this part and for part (d) as well.

g(1) = -2

f( g(1) ) = f(-2) = 5

<h3>Answer: 5</h3>

==========================================================

Part (d)

Same idea as part (c)

g(2) = 0

f( g(2) ) = f( 0 ) = 3

<h3>Answer: 3</h3>

6 0

3 years ago

A standard weight known to weigh 10 grams. Some suspect bias in weights due to manufacturing process. To assess the accuracy of

notsponge [240]

Answer:

a) The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams

b) 49 measurements are needed.

Step-by-step explanation:

Question a:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.98}{2} = 0.01$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.01 = 0.99$ , so Z = 2.327.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 2.327\frac{0.0003}{\sqrt{5}} = 0.00031$

The lower end of the interval is the sample mean subtracted by M. So it is 10.0044 - 0.00031 = 10.00409 grams

The upper end of the interval is the sample mean added to M. So it is 10 + 0.00031 = 10.00471 grams

The 98% confidence interval for the mean weight is between 10.00409 grams and 10.00471 grams.

(b) How many measurements must be averaged to get a margin of error of +/- 0.0001 with 98% confidence?

We have to find n for which M = 0.0001. So

$M = z\frac{\sigma}{\sqrt{n}}$

$0.0001 = 2.327\frac{0.0003}{\sqrt{n}}$

$0.0001\sqrt{n} = 2.327*0.0003$

$\sqrt{n} = \frac{2.327*0.0003}{0.0001}$

$(\sqrt{n})^2 = (\frac{2.327*0.0003}{0.0001})^2$

$n = 48.73$

Rounding up

49 measurements are needed.

7 0

3 years ago

Last time, one out of every 4 students in our class was absent. This time, after Jen came back, only one out of every 5 students

VMariaS [17]

Answer: There are 20 students and 4 students are not here today.

Step-by-step explanation:

If the number of students = 4

Let the number of students be 'x'.

Fraction of students absent = $\dfrac{x}{4}$

If the number of students = 5

Fraction of students absent = $\dfrac{x}{5}$

And Jen came back,

So, the fraction of student absent is also written as

$\dfrac{x}{4}-1$

According to question, it becomes,

$\dfrac{x}{5}=\dfrac{x}{4}-1\\\\\dfrac{x}{5}=\dfrac{x-4}{4}\\\\4x=5(x-4)\\\\4x=5x-20\\\\4x-5x=-20\\\\-x=-20\\\\x=20$

Hence, there are 20 students in our school.

And number of students are not here today is $\dfrac{20}{5}=4$

3 0

3 years ago