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SpyIntel [72]
3 years ago
14

Jacki evaluated the expression below.

Mathematics
2 answers:
lbvjy [14]3 years ago
7 0

Answer:

C Jack did not subtract 12 from 8 correctly

Step-by-step explanation:

2^3(3-1)+4(8-12)=2^3(2)+4(-4)

it should be -4, not +4

Minchanka [31]3 years ago
5 0
Jackie didn’t subtract 12 from 8 right, I think
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A web site was hit 300 times over a period of 15 days Show that over some period of 3 consecutive days, it was hit at least 60 t
marshall27 [118]

Answer:

We will divide the 15 days in five periods of 3 consecutive days each.

Now to solve this we will use the pigeonhole principle.

This states that if (N+1) pigeons occupy N holes, then some hole must have at least 2 pigeons.

So, we have n=300 pigeons  and k=5 holes.

[\frac{n}{k} ]=[\frac{300}{5} ]

Hence, there is a period of 3 consecutive days in which the website was hit at least 60 times.

6 0
3 years ago
Laura and Brent paddled a canoe 6 miles upstream in four hours. The return trip took three hours. Find the rate at which Laura a
mixas84 [53]
Call the rate in still water vs;
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where v is the rate with which the river flows.
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vs+v=2
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5 0
3 years ago
Suppose a researcher is conducting a hypothesis test for a mean and finds a test statistic of z=2.3z=2.3 . Which of the choices
Galina-37 [17]

Answer:

Option c. The result is significant at the 0.020.02 level.

Step-by-step explanation:

We are given the following in the question:

We are conducting a one tailed hypothesis test with alternate hypothesis

H_A: \mu > \mu_0

The z test calculated for the same is

z=2.3

We calculate the p-value from the standard normal table.

p value = 0.010724 at 0.05 significance level.

The result is significant

p value = 0.010724 at 0.01 significance level.

The result is not significant

p value = 0.010724 at 0.02 significance level.

The result is significant at 0.02 significance level.

Thus, the correct answer is

Option c. The result is significant at the 0.020.02 level.

3 0
3 years ago
Ammeters produced by a manufacturer are marketed under the specification that the standard deviation of gauge readings is no lar
AURORKA [14]

Answer:

0.101

Step-by-step explanation:

Given that :

Standard deviation, s = 0.2

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Sample size, n = 10

Population variance = σ² = 0.04

We use the Chi square distribution :

(n - 1)s² / σ²

For P(s² ≥ 0.065)

(n - 1)s² / σ² = (10 - 1)*0.0653 / 0.04

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P(s² ≥ 14.625) = 0.101 ( chi squee calculator).

8 0
2 years ago
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