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3 years ago

What method is this?

Mathematics

2 answers:

aleksandr82 [10.1K]3 years ago

8 0

Answer : nothing
Explaination : “ Nothing Method ” is found by someone in this world

Colt1911 [192]3 years ago

4 0

Answer:

theres nothing

Step-by-step explanation:

so the method is nothing

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In △CDE , DE=15 , CE=7 , and m∠E=38∘ .

Stels [109]

Step-by-step explanation:

thanks for the question!!

I hope it helps you

5 0

3 years ago

Look at the graph below. what is the slope of the line ?

kykrilka [37]

Answer:

The slope is $-\frac{1}{5}$ .

Step-by-step explanation:

Choose 2 points on the graph.

(7,3) and (2,2)

Use slope formula: $\frac{y2-y1}{x2-x1}$

$\frac{2-3}{2-7}$ = $\frac{-1}{-5}$

The slope is $-\frac{1}{5}$ .

8 0

2 years ago

Read 2 more answers

A tin of paint, 6 cm high, holds a half a litre of paint. How much paint would go into a similar

dusya [7]

Answer:

One liter

Step-by-step explanation:

The first tin is 6 cm, holds 1/2 liter.

12 is double 6, because 6 x 2 = 12.

What you do to one side you have to do to the other, 1/2 x 2 = 1.

One liter of paint.

7 0

3 years ago

Randomly selected 110 student cars have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly select

monitta

Answer:

1. Yes, there is sufficient evidence to support the claim that student cars are older than faculty cars.

2. The 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

Step-by-step explanation:

We are given that randomly selected 110 student cars to have ages with a mean of 8 years and a standard deviation of 3.6 years, while randomly selected 75 faculty cars to have ages with a mean of 5.3 years and a standard deviation of 3.7 years.

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

$\mu_2$ = mean age of faculty cars.

So, Null Hypothesis, $H_0$ : $\mu_1 \leq \mu_2$ {means that the student cars are younger than or equal to faculty cars}

Alternate Hypothesis, $H_A$ : $\mu_1>\mu_2$ {means that the student cars are older than faculty cars}

(1) 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, $s_p=\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(110-1)\times 3.6^{2}+(75-1)\times 3.7^{2} }{110+75-2} }$ = 3.641

So, the test statistics = $\frac{(8-5.3)-(0)} {3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} } }$ ~ $t_1_8_3$

= 4.952

The value of t-test statistics is 4.952.

Since the value of our test statistics is more than the critical value of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we support the claim that student cars are older than faculty cars.

(2) The 98% confidence interval for the difference between the two population means ( $\mu_1-\mu_2$ ) is given by;

98% C.I. for ( $\mu_1-\mu_2$ ) = $(\bar X_1-\bar X_2) \pm (t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} })$

= $(8-5.3) \pm (2.326 \times 3.641 \times \sqrt{\frac{1}{110}+\frac{1}{75} })$

= $[2.7 \pm 1.268]$

= [1.432, 3.968]

Here, the critical value of t at a 1% level of significance is 2.326.

Hence, the 98% confidence interval for the difference between the two population means is [1.432 years, 3.968 years].

7 0

3 years ago

What is the value of |−25|? (1 point)

REY [17]

Answer:

25:)

Step-by-step explanation:

4 0

3 years ago