The sum of two consecutive intergers is no more than 209 . Whatbis the large of two integers

Rhea sold 6 boxes of blueberry muffins and 8 boxes of corn muffins at the bake sale. Each box contained 18 muffins. The muffins

Elenna [48]

6 boxes of blueberry muffins and 8 boxes of corn muffins

Since we know that each box has 18 muffins, we can multiply both 6 and 8 to determine the amount of muffins present

6 • 18 = 108 blueberry muffins
8 • 18 = 144 corn muffins

To determine the price, you multiply each type of muffin by 0.50 as each muffin is worth $0.50

108 • 0.50 = 54
144 • 0.50 = 72

You add these two together to get the price

54 + 72 = $126

If you explain what the estimate is, if I didn’t answer that along the way, I can help answer it for you.

7 0

2 years ago

I feel like I am asking so my questions but I dont really care its a Sunday xD

dimulka [17.4K]

You would add like terms so it would be 1x then you add 1 to 7 so you would get 8.. the you divide both sides by 1x and you'd get 8=x

4 0

3 years ago

Read 2 more answers

Can someone pls help, I'll answer your questions if you have any.

Pie

There are no pictures I see you can text me on insta if you need help original_loredo

3 0

3 years ago

On a particular production line,the likelihood that a light bulb is defective is 5%. Ten light bulbs are randomly selected. What

KengaRu [80]

Answer:

Mean and Variance of the number of defective bulbs are 0.5 and 0.475 respectively.

Step-by-step explanation:

Consider the provided information,

Let X is the number of defective bulbs.

Ten light bulbs are randomly selected.

The likelihood that a light bulb is defective is 5%.

Therefore sample size is = n = 10

Probability of a defective bulb = p = 0.05.

Therefore, q = 1 - p = 1 - 0.05 = 0.95

Mean of binomial random variable: $\mu=np$

Therefore, $\mu=10(0.05)=0.5$

Variance of binomial random variable: $\sigma^2=npq$

Therefore, $\sigma^2=10(0.05)(0.95)=0.475$

Mean and Variance of the number of defective bulbs are 0.5 and 0.475 respectively.

5 0

3 years ago

A psychological experiment was conducted to investigate the length of time (time delay) between the administration of a stimulus

aleksklad [387]

Answer:

The P-value of this sample is 0 and is less than the significance level (0.05), so the effect is significant and the null hypothesis is rejected.

As there is significant evidence to reject $H_0: \mu= 1.6$ , we can say that there is significant evidence to claim that the mean time delay for the hypothetical population of all persons who may be subjected to the stimulus differs from 1.6 seconds.

The significance level for this test is 0.05.

Step-by-step explanation:

In this case, we want to prove if there is significant evidence that the mean differs from 1.6 seconds. That is the same as having evidence to reject the the hypothesis $H_0:\mu= 1.6$ (the null hypothesis always have the equal sign).

We have to test the null hypothesis

$H_0:\mu= 1.6$

The significance level for this test is 0.05

Calculation of the t-statistic:

$t=\frac{M-\mu}{s/\sqrt{n}} =\frac{2.2-1.6}{0.57/\sqrt{36}} =\frac{0.6}{0.063}= 9.47$

If we look up in a t-table for t=9.47 and df=(36-1)=35, we get this value appears with a probability of zero. A large number like that is very unlikely to happen.

The P-value of this sample is 0 and is less than the significance level (0.05), so the effect is significant and the null hypothesis is rejected.

As there is significant evidence to reject $H_0: \mu= 1.6$ , we can say that there is significant evidence to claim that the mean time delay for the hypothetical population of all persons who may be subjected to the stimulus differs from 1.6 seconds.

6 0

3 years ago