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Nookie1986 [14]

3 years ago

8

Max is going to eat either a slice of pie, an apple, or drink a cup of cider for a snack. He is also going to buy pumpkin or a h

at. How many different ways can max eat one snack or buy one item

2 answers:

MAXImum [283]3 years ago

6 0

Answer:

6 different ways

Step-by-step explanation:

pie, pumpkin

pie, hat

apple, pumpkin

apple, hat

cider, pumpkin

cider, hat

sattari [20]3 years ago

4 0

6 different ways! ....

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Please help me this is important

Dmitrij [34]

Answer:

C. II, III, and IV only

Step-by-step explanation:

Lol I haven't done one of these in a while, so thanks for the practice!

It's important to note that a dilation does NOT change the angles of the shape whatsoever. By this, you know that II and III are automatically correct, which rules out A and B. After that, you can look at I, which is the only difference between C and D. Since the dilation factor is 2/3, the parallelogram you see is smaller than the original. Since in I, the original would be larger than the denominator, the correct answer would be at least greater than one (to be precise, it'd be 3/2). To make everything easier, you can find the value by assigning the original side a value of 1, which would make the prime value 2/3. 1/(2/3) is going to be 3/2.

4 0

3 years ago

Does anyone understand this

melomori [17]

I hope this helps you

7 0

3 years ago

Find an expression which represents the sum of (-8x+7y)(−8x+7y) and (2x-2y)(2x−2y) in simplest terms.

Yakvenalex [24]

Answer:

(-8x+7y)(−8x+7y)+(2x-2y)(2x−2y)= simplified - 68x^2+53y^2−120xy

I hope this helps ! <3

Step-by-step explanation:

3 0

3 years ago

A genetic experiment involving peas yielded one sample of offspring consisting of 420 green peas and 174 yellow peas. Use a 0.01

slavikrds [6]

Answer:

a) $z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

b) For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

Step-by-step explanation:

Data given and notation

n=420+174=594 represent the random sample taken

X=174 represent the number of yellow peas

$\hat p=\frac{174}{594}=0.293$ estimated proportion of yellow peas

$p_o=0.23$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

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 of yellow peas is 0.23:

Null hypothesis: $p=0.23$

Alternative hypothesis: $p \neq 0.23$

When we conduct a proportion test we need to use the z statisitc, 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:

$z=\frac{0.293 -0.23}{\sqrt{\frac{0.23(1-0.23)}{594}}}=3.649$

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 bilateral test the p value would be:

$p_v =2*P(z>3.649)=0.00026$

So the p value obtained was a very low value 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.

b) Critical value

For this case we need to find a critical value that accumulates $\alpha/2$ of the area on each tail, we know that $\alpha=0.01$ , so then $\alpha/2 =0.005$ , using the normal standard table or excel we see that:

$z_{crit}= \pm 2.58$

Since the calculated value is higher than the critical value we have enough evidence to reject the null hypothesis at 1% of significance.

5 0

3 years ago

How many times will 1/5 go into 37?

Afina-wow [57]

Answer:

185 times

Step-by-step explanation:

5 0

3 years ago

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