Which expression are equivalent to 6+12x

Can you help me with number 2 please show me how you did it thank you

denpristay [2]

Distraibutiv peorpoty
a(b+c)=ab+ac

we have
3a+7+2(3+a)
distribute first
2(3+a)=6+2a

now we have
3a+7+6+2a
move like terms together
3a+2a+7+6
add lke terms
5a+13

7 0

3 years ago

Which proportion could be used to determine if the figures represent a dilation? 2/3=5/8, 2/3=8/5, 2/8=5/3, or 2/8=3/5?

Maurinko [17]

Answer:

The answer is A. 2/3 =5/8

Step-by-step explanation:

5 0

3 years ago

A ballpark sells bags of popcorn for $1 each and bags of peanuts for $2 each. Hector will buy x bags of popcorn and y bags of pe

Rasek [7]

Answer:

Hector will buy 6 peanut bags and he will buy 2 ppopcorn bags

Step-by-step explanation:

He will buy 6 bags of peanuts because that will be 12$ and he will buy 2 popcorn bags because it will be 2$. He only wanted to buy 8 bags of snacks and he only wanted to spend 14$. Hope this helps!!! :)

6 0

3 years ago

Solve<br> x + y = 100 x -y = 10<br> Pair of Equation

Kazeer [188]

lets take

x+y=100 firstly.

then x=100-y equation (i)

Now,

x-y=10

Putting value of x from equation(i)

100-y-y=10

-2y=10-100

-2y=-90

therefore, y=45

Now,

putting value of y in equation(i)

we get

x=100-45

ie x=55

hence, x=55 & y=45

8 0

3 years ago

Read 2 more answers

A random sample of 100 people was taken. Eighty-five of the people in the sample favored Candidate A. We are interested in deter

lbvjy [14]

Answer:

$z=\frac{0.85 -0.8}{\sqrt{\frac{0.8(1-0.8)}{100}}}=1.25$

$p_v =P(z>1.25)=0.106$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who favored candidate A is not significalty higher than 0.8

ii) not significantly greater than 80%

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

$\hat p=0.85$ estimated proportion of people who favored candidate A

$p_o=0.8$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the true proportion is higher than 0.8.:

Null hypothesis: $p\leq 0.8$

Alternative hypothesis: $p > 0.8$

When we conduct a proportion test we need to use the z statistic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.85 -0.8}{\sqrt{\frac{0.8(1-0.8)}{100}}}=1.25$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.05$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>1.25)=0.106$

If we compare the p value obtained and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of people who favored candidate A is not significalty higher than 0.8

ii) not significantly greater than 80%

8 0

3 years ago