What is the x intercept of a line that passed through the point (3,4) an has a slope of 2

PLEASE HELP!! Multiply and Simplify 10x^2/6y^3 * 24y^2/35x^2

Llana [10]

Your answer is: 8x^4 y^5/7.

7 0

4 years ago

Change the width 20/16 inches to a mixed numeral in simplest form.

Iteru [2.4K]

Answer:1 and 1/4

Step-by-step explanation:

Okay, let's see the given: 20/16

So, we can take out 16 from the top to make it 1 4/16

Now, we can simplify that by dividing the top and bottom by 2: 1 2/8

We can simplify that again to get 1 1/4

So the answer is 1 and 1/4

5 0

3 years ago

What is the solution of the system.. explain what it means in the solution

UNO [17]

Answer:

The solution is (4,250)

Step-by-step explanation:

This means that when a member of either gym (A or B) signs up for a four-month membership, they pay the same amount - $250. Hope this helps :)

5 0

3 years ago

Which technique is most appropriate to use to solve each equation? Drag the name of the technique into the box to match each equ

NemiM [27]

For the first equation, the answer is C) completing the square.
For the second equation, the answer is B) zero product property.

For the first equation, we can easily complete the square by finding half of b and squaring it; then we can take the square root of both sides and solve the equation.

For the second equation, since it is already factored, we use the zero product property to solve it.

7 0

4 years ago

Suppose that 70% of college women have been on a diet within the past 12 months. A sample survey interviews an SRS of 267 colleg

Fofino [41]

Answer:

The probability that 75% or more of the women in the sample have been on a diet is 0.037.

Step-by-step explanation:

Let X = number of college women on a diet.

The probability of a woman being on diet is, P (X) = p = 0.70.

The sample of women selected is, n = 267.

The random variable thus follows a Binomial distribution with parameters n = 267 and p = 0.70.

As the sample size is large (n > 30), according to the Central limit theorem the sampling distribution of sample proportions ( $\hat p$ ) follows a Normal distribution.

The mean of this distribution is:

$\mu_{\hat p} = p = 0.70$

The standard deviation of this distribution is: $\sigma_{\hat p}=\sqrt{\frac{p(1-p}{n}} =\sqrt{\frac{0.70(1-0.70}{267}}=0.028$

Compute the probability that 75% or more of the women in the sample have been on a diet as follows:

$P(\hat p \geq 0.75)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}} \geq \frac{0.75-0.70}{0.028}) =P(Z\geq0.179)=1-P(Z$

**Use the z-table for the probability.

$P(\hat p \geq 0.75)=1-P(Z$

Thus, the probability that 75% or more of the women in the sample have been on a diet is 0.037.

7 0

3 years ago