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3 years ago

How do i solve for x with mixture word problems

1 answer:

4 0

Well, solving X works different ways in different situations. Say the problem is 3x=6 you have to figure out what to multiply to get the end answer. Now, say it is 2+x=8 what can you add to 2 in order to get 8? Another way is if it says 8+2=x you have to figure out what 8 plus 2 is and that is what X will be. I really hope this helps! :)

You might be interested in

2) BRAINLIEST + 10 + POINTS :D

elena55 [62]

Answer:

c) 28

explanation :

In this sequence:

a1 = 2

a2 = 4

a3 = 6

a4 = 8

a5 = 10

and so on...

then

the sum from a2 to a5 is:

a2 +a3 + a4 + a5 = 4 + 6 +8 + 10 = 28

7 0

3 years ago

A statistics practitioner in a large university is investigating the factors that affect salary of professors. He wondered if ev

Ulleksa [173]

Answer:

Step-by-step explanation:

Hello!

Given the linear regression of Y: "Annual salary" as a function of X: "Mean score on teaching evaluation" of a population of university professors. It is desired to study whether student evaluations are related to salaries.

The population equation line is

E(Y)= β₀ + β₁X

Using the information of a n= 100 sample, the following data was calculated:

R²= 0.23

Coefficient Standard Error

Intercept 25675.5 11393

x 5321 2119

The estimated equation is

^Y= 25675.5 + 5321X

Now if the interest is to test if the teaching evaluation affects the proffesor's annual salary, the hypotheses are:

H₀: β = 0

H₁: β ≠ 0

There are two statistic you can use to make this test, a Student's t or an ANOVA F.

Since you have information about the estimation of β you can calculate the two tailed t test using the formula:

$t= \frac{b - \beta }{\frac{Sb}{\sqrt{n} } }$ ~ $t_{n-2}$

$t= \frac{ 5321 - 0 }{\frac{2119}{\sqrt{100} } }$ = 25.1109

The p-value is two-tailed, and is the probability of getting a value as extreme as the calculated $t_{H_0}$ under the distribution $t_{98}$

p-value < 0.00001

I hope it helps!

3 0

3 years ago

1. Find the mean of the data set: 11, 8, 18, 9, 15, 11. Please help asap!

Stels [109]

Answer:

Step-by-step explanation:

The mean is the average. Which is all of the numbers added together divided by the amount of numbers.

11+8+18+9+15+11=72

72÷6=12

7 0

3 years ago

Read 2 more answers

The Oregon Department of Health web site provides information on the cost-to-charge ratio (the percentage of billed charges that

Alekssandra [29.7K]

Answer:

We conclude that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care.

Step-by-step explanation:W

We are given with the cost-to-charge ratios for both inpatient and outpatient care in 2002 for a sample of six hospitals in Oregon below;

Hospital 2002 Inpatient Ratio 2002 Outpatient Ratio

1 68 54

2 100 75

3 71 53

4 74 56

5 100 74

6 83 71

Let $\mu_1$ = mean cost-to-charge ratio for outpatient care

$\mu_2$ = mean cost-to-charge ratio for impatient care.

SO, Null Hypothesis, $H_0$ : $\mu_1 \geq \mu_2$ {means that the mean cost-to-charge ratio for Oregon hospitals is higher or equal for outpatient care than for inpatient care}

Alternate Hypothesis, $H_A$ : $\mu_1 < \mu_2$ {means that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care}

The test statistics that would be used here is Two-sample t-test statistics because we don't know about 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 cost-to-charge Outpatient Ratio = $\frac{\sum X_1}{n_1}$ = 63.83

$\bar X_2$ = sample mean cost-to-charge Impatient Ratio = $\frac{\sum X_2}{n_2}$ = 82.67

$s_1$ = sample standard deviation for Outpatient Ratio = $\sqrt{\frac{\sum (X_1-\bar X_1 )^{2} }{n_1-1} }$ = 10.53

$s_2$ = sample standard deviation for Impatient Ratio = $\sqrt{\frac{\sum (X_2-\bar X_2 )^{2} }{n_2-1} }$ = 14.33

$n_1$ = sample of hospital for outpatient care = 6

$n_2$ = sample of hospital for outpatient care = 6

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(6-1)\times 10.53^{2}+(6-1)\times 14.33^{2} }{6+6-2} }$ = 12.574

So, the test statistics = $\frac{(63.83-82.67)-(0)}{12.574 \times \sqrt{\frac{1}{6}+\frac{1}{6} } }$ ~ $t_1_0$

= -2.595

The value of t test statistics is -2.595.

Now, at 5% significance level, the t table gives critical value of -1.812 at 10 degree of freedom for left-tailed test.

Since, our test statistics is less than the critical value of t as -2.595 < -1.812, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean cost-to-charge ratio for Oregon hospitals is lower for outpatient care than for inpatient care.

8 0

3 years ago

Solve the polynomial congruence x² ≡ 1(mod3).

igor_vitrenko [27]

Answer:

x is any integer not a multiple of 3

Step-by-step explanation:

This is equivalent to ...

(x mod 3)² = 1

x mod 3 = ±1 . . . . . . taking the square root

Possible values of x mod 3 are 0, 1, 2 (≅-1), so the only value excluded is 0.

x mod 3 ≠ 0 . . . . . . for any integer x

3 0

3 years ago