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Darya [45]
3 years ago
13

Four students estimated the width of the door to their classroom. Who made the best estimate?

Mathematics
1 answer:
VikaD [51]3 years ago
8 0
It depends on what the estimates were
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Evaluate e – 2 when e=7.
Readme [11.4K]

Answer:

5

Step-by-step explanation:

If <em>e = 7,</em> you would replace e in the expression <em>e - 2 </em>with 7 so that the expression becomes <em>7 - 2</em> which results in 5.

Does that help?

3 0
3 years ago
Read 2 more answers
Of 150 adults selected randomly from one town, 30 of them smoke. Construct a 99% confidence interval for the true percentage of
anyanavicka [17]

Answer:

The 99% confidence interval for the true percentage of all adults in the town that smoke is (0.1159, 0.2841).

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:

Of 150 adults selected randomly from one town, 30 of them smoke. This means that n = 150, p = \frac{30}{150} = 0.2

99% confidence interval

So \alpha = 0.01, z is the value of Z that has a pvalue of 1 - \frac{0.01}{2} = 0.995, so Z = 2.575.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.20 - 2.575\sqrt{\frac{0.20*0.80}{150}} = 0.1159

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.20 + 2.575\sqrt{\frac{0.20*0.80}{150}}{119}} = 0.2841

The 99% confidence interval for the true percentage of all adults in the town that smoke is (0.1159, 0.2841).

3 0
3 years ago
If a x b express 12x(10x5)in simplest form
Mrac [35]
That's easy but I'll give the answer is 600
5 0
3 years ago
g Suppose a factory production line uses 3 machines, A, B, and C for making bolts. The total output from the line is distributed
Vladimir [108]

Answer:

The probability that it came from A, given that is defective is 0.362.

Step-by-step explanation:

Define the events:

A: The element comes from A.

B: The element comes from B.

C: The element comes from C.

D: The elemens is defective.

We are given that P(A) = 0.25, P(B) = 0.35, P(C) = 0.4. Recall that since the element comes from only one of the machines, if an element is defective, it comes either from A, B or C. Using the probability axioms, we can calculate that

P(D) = P(A\cap D) + P(B\cap D) + P(C\cap D)

Recall that given events E,F the conditional probability of E given F is defined as

P(E|F) = \frac{P(E\cap F)}{P(F)}, from where we deduce that

P(E\cap F) = P(E|F)P(F).

We are given that given that the element is from A, the probability of being defective is 5%. That is P(D|A) =0.05. Using the previous analysis we get that

P(D) = P(A)P(D|A)+P(B) P(D|B) + P(C)P(D|C) = 0.05\cdot 0.25+0.04\cdot 0.35+0.02\cdot 0.4 = 0.0345

We are told to calculate P(A|D), then using the formulas we have

P(A|D) = \frac{P(A\cap D)}{P(D)}= \frac{P(D|A)P(A)}{P(D)}= \frac{0.05\cdot 0.25}{0.0345}= 0.36231884

3 0
2 years ago
11/1,000 as a decimal
xz_007 [3.2K]
Divide your numerator by the denominator to find the decimal. 11÷1000=.011 ANSWER: .011
5 0
3 years ago
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