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2 years ago

Please answer this correctly without making mistakes

Mathematics

1 answer:

Dennis_Churaev [7]2 years ago

3 0

9514 1404 393

Answer:

35 gallons 0 quarts

Step-by-step explanation:

There are 4 quarts per gallon, so this is ...

1 gallon 1 quart = 1 1/4 gallons

Then the product is ...

(1 1/4 gallons) × 28 = (5/4 gallons)×28 = 35 gallons (and 0 quarts)

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All outcomes being equally likely, what is the probability a classmate receives a gift bag with a bouncy ball?

nekit [7.7K]

I think the answer is b and c but I’m not very sure

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2 years ago

Factor n^2+5n-24. Show that the original polynomial and the factored from describe the same sequence of numbers for n=0,1,2,3,an

Lana71 [14]

Step-by-step explanation:

We have the original equation n ^ 2 + 5 * n - 24, when factoring we have:

(n + 8) * (n - 3)

Now by replacing the values:

n = 0

0 ^ 2 + 5 * 0 - 24 = - 24

(0 + 8) * (0 - 3) = - 24

n = 1

1 ^ 2 + 1 * 0 - 24 = - 18

(1 + 8) * (1 - 3) = -18

n = 2

2 ^ 2 + 5 * 2 - 24 = - 10

(2 + 8) * (2 - 3) = - 10

n = 3

3 ^ 2 + 5 * 3 - 24 = 0

(3 + 8) * (3 - 3) = 0

n = 4

4 ^ 2 + 5 * 4 - 24 = 12

(4 + 8) * (4 - 3) = 12

7 0

2 years ago

What properties of equality do you need to use to solve the equation 5.6x−16.9=−0.1? Solve the equation

kakasveta [241]

Answer:

5.6x-16.9=-0.1

28x/5-16.9=-0.1

28x/5-16.9+16.9=-0.1+16.9

28x/5=16.8

x=3

Step-by-step explanation:

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2 years ago

Question

ivann1987 [24]

Answer:

$195.30

Step-by-step explanation:

$180 + 8.5 = $195.30

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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