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2.4 liters of water is poured into a pitcher that now contains 10.1 liters of water. Which equation represents this?

Alenkinab [10]

I couldn’t figure out d answer to This hope this helped

8 0

3 years ago

The question is attached at the bottom. PLEASE HELP with detailed explanation!!!

Eva8 [605]

The correct answer are 1 , 3 , 4

8 0

3 years ago

I need help with geometry plz

Softa [21]

Answer:

Point A is -2, 1

Point B is -4, -2

Point C is -4, -6

and D is -1, 3.

Step-by-step explanation:

Irdk if this is all right bcz its kinda hard to see the exact points. Hope it is tho, and hope it helps you out! Good luck :P

4 0

3 years ago

The American Bankers Association reported that, in a sample of 120 consumer purchases in France, 48 were made with cash, compare

tatuchka [14]

Answer:

Step-by-step explanation:

Hello!

You have the information for two variables

X₁: Number of consumer purchases in France that were made with cash, in a sample of 120.

n₁= 120 consumer purchases

x₁= 48 cash purchases

p'₁= 48/120= 0.4

X₂: Number of consumer purchases in the US that were made with cash, in a sample of 55.

n₂= 55 consumer purchases

x₂= 24 cash purchases

p'₂= 24/55= 0.4364

You need to construct a 90% CI for the difference of proportions p₁-p₂

Using the central limit theorem you can approximate the distribution of both sample proportions p'₁ and p'₂ to normal, so the statistic to use to estimate the difference of proportions is an approximate standard normal:

[(p'₁-p'₂) ± $Z_{1-\alpha /2}$ * $\sqrt{\frac{p'_1(1-p'_1)}{n_1} +\frac{p'_2(1-p'_2)}{n_2} }$ ]

$Z_{0.95}= 1.648$

[(0.4-0.4364)±1.648 * $\sqrt{\frac{0.4(1-0.4)}{120} +\frac{0.4364(1-0.4364)}{55} }$ ]

[-0.1689;0.0961]

The interval has a negative bond, it is ok, keep in mind that even tough proportions take values between 0 and 1, in this case, the confidence interval estimates the difference between the two proportions. It is valid for one of the bonds or the two bonds of the CI for the difference between population proportions to be negative.

I hope this helps!

8 0

4 years ago

The fraction of defective integrated circuits produced in a photolithography process is being studied. A random sample of 300 ci

Olenka [21]

Answer:

The correct answer is

(0.0128, 0.0532)

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence interval $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$

For this problem, we have that:

In a random sample of 300 circuits, 10 are defective. This means that $n = 300$ and $\pi = \frac{10}{300} = 0.033$

Calculate a 95% two-sided confidence interval on the fraction of defective circuits produced by this particular tool.

So $\alpha$ = 0.05, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{300}} = 0.033 - 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0128$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{300}} = 0.033 + 1.96\sqrt{\frac{0.033*0.967}{300}} = 0.0532$

The correct answer is

(0.0128, 0.0532)

4 0

3 years ago