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3 years ago

Photos are attached, large amount of points reward, please answer correctly, will give brainliest!

Mathematics

1 answer:

kompoz [17]3 years ago

4 0

I can’t answer part c but

The x-int = (24,0)
The y-int = (0,20)
The slope = -5/6

Here’s what the graph would look like

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Explain how you are able to get the answer without having to multiply 1.23 by 10 thirteen times.

max2010maxim [7]

We need more context to answer your question

5 0

2 years ago

A rectangle has perimeter 182 inches and length 52 inches what is the width

zimovet [89]

If the perimeter is 182, double the length which is 52 to get 104. Subtract that from 182 to get 78. Divide that by 2 to get the width which is 39.

4 0

3 years ago

Read 2 more answers

Find the 12th term of the geometric sequence 8, 16, 32...

Yuki888 [10]

Answer:

a₁₂ = 16384

Step-by-step explanation:

The nth term of a geometric sequence is

$a_{n}$ = a₁ $(r)^{n-1}$

where a₁ is the first term and r the common ratio

Here a₁ = 8 and r = a₂ ÷ a₁ = 16 ÷ 8 = 2 , then

a₁₂ = 8 × $2^{11}$ = 8 × 2048 = 16384

3 0

3 years ago

Read 2 more answers

Geometry: The rectangle shown has a perimeter of 34cm and the given area. Its length is 5 more than twice it's width. Write and

Thepotemich [5.8K]

w - width

2w + 5 - length

w + w + (2w + 5) + (2w + 5) = 6w + 10 - perimeter

34 cm - perimeter

The equation:

6w + 10 = 34 subtract 10 from both sides

6w = 24 divide both sides by 6

w = 4 cm

2w + 5 = 2(4) + 5 = 8 + 5 = 13 cm

<h3>Answer: Width = 4cm, Length = 13cm</h3>

3 0

2 years ago

Two statistics teachers both believe that each has the smarter class. To put this to the test, they give the same final exam to

SVETLANKA909090 [29]

Answer:

We conclude that there is no difference between the two classes.

Step-by-step explanation:

We are given that two statistics teachers both believe that each has a smarter class.

A summary of the class sizes, class means, and standard deviations is given below:n1 = 47, x-bar1 = 84.4, s1 = 18n2 = 50, x-bar2 = 82.9, s2 = 17

Let $\mu_1$ = mean age of student cars.

$\mu_2$ = mean age of faculty cars.

So, Null Hypothesis, $H_0$ : $\mu_1=\mu_2$ {means that there is no difference in the two classes}

Alternate Hypothesis, $H_A$ : $\mu_1\neq \mu_2$ {means that there is a difference in the two classes}

The test statistics that will be used here is Two-sample t-test statistics because we don't know about the population standard deviations;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean age of student cars = 8 years

$\bar X_2$ = sample mean age of faculty cars = 5.3 years

$s_1$ = sample standard deviation of student cars = 3.6 years

$s_2$ = sample standard deviation of student cars = 3.7 years

$n_1$ = sample of student cars = 110

$n_2$ = sample of faculty cars = 75

Also, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(47-1)\times 18^{2}+(50-1)\times 17^{2} }{47+50-2} }$ = 17.491

So, the test statistics = $\frac{(84.4-82.9)-(0)}{17.491 \times \sqrt{\frac{1}{47}+\frac{1}{50} } }$ ~ $t_9_5$

= 0.422

The value of t-test statistics is 0.422.

Now, the P-value of the test statistics is given by;

P-value = P( $t_9_5$ > 0.422) = From the t table it is clear that the P-value will lie somewhere between 40% and 30%.

Since the P-value of our test statistics is way more than the level of significance of 0.04, so we have insufficient evidence to reject our null hypothesis as our test statistics will not fall in the rejection region.

Therefore, we conclude that there is no difference between the two classes.

6 0

2 years ago