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Mrrafil [7]
3 years ago
6

Help! I need somebody

Mathematics
1 answer:
oee [108]3 years ago
6 0

Answer:

t to the sixth power equals 9

t=1.5

Step-by-step explanation:

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So I have 4 math problems, whether you choose to answer them or not is your choice. Also thank you if you do answer them. :)
cestrela7 [59]

Answer:

1) 9

2) 256

3) -32

4) -2744

Step-by-step explanation:

3 0
3 years ago
A water cooler holds 15 liters of sports drink. Approximately how many gallons is this? (1 gallon = 3.785 liters)
Katena32 [7]
Approximately 3.9 gallons I’m sure of
4 0
3 years ago
Which statement implies that A and B are independent events?
STatiana [176]

Answer:

P(A \cap B) = P(A)P(B)

Step-by-step explanation:

Independent events:

If two events, A and B are independent, the probability of both A and B happening is the same as the probability of A happening multiplied by the probability of B happenings, that is:

P(A \cap B) = P(A)P(B)

In this question:

The statement is P(A \cap B) = P(A)P(B)

4 0
3 years ago
The researchers are concerned that the dollars spent per shopper examined in the study is too different from what has been found
Katarina [22]

Answer:

Step-by-step explanation:

Hello!

To study if the average dollars spend per shopper is different than the expected population average of $84.5 there was a sample of 120 shoppers taken and their spending habit registered.

Then the study variable is:

X: Amount of money spent by a shopper ($)

This population standard deviation is σ= $16

Using the data information I've run a normality test, with p-value: 0.3056 against the test significance level α: 0.05, the decision is to not reject the null hypothesis so there is enough statistical evidence to conclude that the study variable has a normal distribution.

X~N(μ;σ²)

I've also calculated the sample mean and standard deviation:

X[bar]= $81.01

S= $19.62

a.

If the objective is to test that the true mean spending for the population is significantly different than the expected average of 84.5, the parameter of interest is the population mean μ and the hypothesis test will be two-tailed.

H₀: μ = 84.5

H₁: μ ≠ 84.5

α:0.05

The statistic is a standard normal for one sample:

Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~N(0;1)

Z_{H_0}= \frac{81.01-84.5}{\frac{16}{\sqrt{120} } } = -2.389

p-value two tailed  0.016894

b.

The decision rule using the p-value method is:

If p-value ≤ α ⇒ Reject the null hypothesis.

If p-value > α ⇒ Do not reject the null hypothesis.

Since the p-value is less than the significance level, the decision is to reject the null hypothesis.

c.

Using a level of significance of 5% the null hypothesis was rejected. So there is enough statistical evidence to support the researcher's claim that the true mean spending for the population is significantly different than the expected average of $84.5.

I hope it helps!

3 0
3 years ago
Which value of x is in the domain of f(x) = square root of x-5
Gelneren [198K]
The answer is A. x= 5
3 0
3 years ago
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