What is -3\4 + (-3\8)

HELPPO<br><br> What are the zeros of the function graphed below?

Leya [2.2K]

It's just wherever the function touches the x axis, so 5 and -1

7 0

3 years ago

The expression 2( 21x + 75 ) represents the monthly tuition and the one-time materials fee that Jamie pays for his two children

ss7ja [257]

Answer: x represents the number of months, 150 represents one-time materials fee, 2( 21x + 75 ) represents total fees in x months and 42 represents rate of change.

Step-by-step explanation:

Since, Here the expression 2( 21x + 75 ) represents the monthly tuition and the one-time materials fee that Jamie pays for his two children to join a science club.

Let, f(x) represents the monthly tuition fee

Therefore, f(x)= 2( 21x + 75 )

⇒f(x)= 42 x +150

Again according to the question, there is a one-time materials fee which must be free from the variable.

Therefore In f(x) , 150 must represent one time material fees.

And, x is variable, and here number of months is a variable.

Thus, x must be denote the number of months.

And, 42 must be denote the fees for 1 month.

8 0

3 years ago

HELP!! ILL MARK U BRAINLIEST!!!

vagabundo [1.1K]

Answer:

6.96

Step-by-step explanation:

4 0

3 years ago

Help me with my math homework

Leno4ka [110]

1 yr = 12 months

- 2931/12 = - 244.25 per month

3 0

3 years ago

A new sales training program has been instituted at a rent-to-own company. Prior to the training, 10 employees were tested on th

Lunna [17]

Answer:

C. H0: µD = 0, HA: µD <0

Step-by-step explanation:

We assume that the paired sample data are simple random samples and that the differences have a distribution that is approximately normal. So for this case is better apply a paired t test.

A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.

Let put some notation

x=test value for pretest , y = test value for posttest

x: 66, 94, 87, 84, 76, 88

y: 75, 100, 93, 85, 75, 90

The system of hypothesis for this case are:

Null hypothesis: $\mu_D \geq 0$

Alternative hypothesis: $\mu_D < 0$

Because the difference D is defined as Pretest-Postest. And we want to see if the postets score is higher than the pretest with th training.

The first step is calculate the difference $d_i=x_i-y_i$ and we obtain this:

d: -9, -6, -6, -1, 1, -2

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{-23}{6}=-3.833$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =3.763$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-3.833 -0}{\frac{3.763}{\sqrt{6}}}=-2.494$

The next step is calculate the degrees of freedom given by:

$df=n-1=6-1=5$

Now we can calculate the p value, since we have a left tailed test the p value is given by:

$p_v =P(t_{(5)}$

The p value is lower than the significance level assumed $\alpha=0.05$ , so then we can conclude that we can reject the null hypothesis. So we can say that the differences between Pretest and Posttest $\mu_{pretest}-\mu_{postest}$ are less than 0.

3 0

3 years ago