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3 years ago

Solve the equations by combining like terms. Show all the steps: 16x − 4x = -48

Mathematics

2 answers:

shusha [124]3 years ago

6 0

Do the Xs have to be the same number

Alexxx [7]3 years ago

5 0

You can combine like/similar terms if the number has the same x at the end, or if the like terms are all regular numbers.
Since 16x and -4x both have an x, you can combine them like you usually would.
16x - 4x = 12x
12x = -48
If you want to solve the equation, you’d divide both numbers by 12 to get your singular x value.
x = -4

Hope that helped! :)

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3 years ago

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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