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Kelly joins Powerhouse Gym. The cost is $30 per month and a one-time $100 joining fee. Lisa joins MegaMuscle Gym for $35 per mon

xeze [42]

In 10 months Lisa will have paid the same amount as Kelly, but in month 11 she will have paid more.

6 0

3 years ago

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What is the domain of the function f (x) x-5/x^2-9

nikklg [1K]

Means the number you can use
remember we can't divide by zero
domain is x^2-9
set to zero to find values we can't use
x^2-9=0
x^2=9
x=+/-3

domain=all real numbers except -3 and 3

7 0

3 years ago

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Question 12 (1 point)

Paraphin [41]

<h3>Answer: 5</h3>

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Explanation:

Let's consider the expression (x-y)^2. It expands out to x^2-2xy+y^2. The terms are:

x^2
-2xy
y^2

Each of those terms either has a single variable with an exponent of 2, or has the exponents add to 2. Think of 2xy as 2x^1y^1.

In short, this means that the degree of each monomial term is 2.

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Now consider (x-y)^3. It expands out into x^3-3x^2y+3xy^2+y^3.

We have terms that either have a single variable and the exponent is 3, or the exponents add to 3. The degree of each term is 3.

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This pattern continues.

In general, for (x-y)^n, where n is any positive whole number, the degree of each term in the expansion is n. If you picked any term, added the exponents, then the exponents will add to n.

4 0

2 years ago

Petrolyn motor oil is a combination of natural oil and synthetic oil. It contains 7 liters of natural oil for every 5 liters of

Svetach [21]

What we know:
Petrolyn motor oil is a combination of natural oil and synthetic oil. It contains 7 liters of natural oil for every 5 liters of synthetic oil.

What we need to find:
If 355 liters of synthetic oil are used, how many liters of petrolyn oil will be made?

Set up a proportion to find how much natural oil is used.
7L natural oil/5L synthetic oil = x/355 synthetic oil where x=natural oil
7/5=x/355
(5)(x)=(7)(355) cross multiplied
5x=2485 simplified
5/5x=2485/5 multiplicative inverse
x=497L

Now, add both synthetic oil and natural oil to find amount of petrolyn oil.

355 L synthetic oil + 497 L natural oil= 852 L of petrolyn oil

5 0

3 years ago

A dataset lists full IQ scores for a random sample of subjects with low lead levels in their blood (sample 1) and another random

Alenkasestr [34]

Answer:

a. Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 >\mu_2$

b. $t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282$

c. $p_v =P(t_{97}>2.287) =0.0122$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).

Step-by-step explanation:

a. State and label the null and alternative hypotheses.

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 >\mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 \leq 0$

Alternative hypothesis: $\mu_1 -\mu_2>0$

Our notation on this case :

$n_1 =78$ represent the sample size for group 1

$n_2 =21$ represent the sample size for group 2

$\bar X_1 =92.88$ represent the sample mean for the group 1

$\bar X_2 =86.90$ represent the sample mean for the group 2

$s_1=15.34$ represent the sample standard deviation for group 1

$s_2=8.99$ represent the sample standard deviation for group 2

b. State the value of the test statistic.

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{s^2_2}{n_2}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom. If we replace the values given we have:

$t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282$

Now we can calculate the degrees of freedom given by:

$df=78+21-2=97$

c. Find either the critical value(s) and draw a picture of the critical region(s) or find the P-value for this test. Indicate which method you are using: ( CIRCLE ONE: Critical value / P-value )

Method used: P value

And now we can calculate the p value using the altenative hypothesis, since it's a right tail test the p value is given by:

$p_v =P(t_{97}>2.287) =0.0122$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).

6 0

3 years ago