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aleksandrvk [35]
2 years ago
7

Guys, I need help ASAP!!

Mathematics
2 answers:
solniwko [45]2 years ago
8 0

Answer:

Pr (3.88 ≤ X ≤ 5.12)

Step-by-step explanation:

the other person is right. i took quizz and got 100% so it has to be the answer

Zarrin [17]2 years ago
7 0

Answer:

The answer is

Pr (3.88 ≤ X ≤ 5.12)

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Which number sentence is true? A. 3.4 x (-3.4) = 11.56 B. 3.4 x 3.4 = 11.56 C. -3.4 x (-3.4) = -11.56 D. -3.4 x 3.4 = 11.56
avanturin [10]

answer b is true 3.4x3.4 = 11.56

3 0
2 years ago
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A mail order company charges 4.5% for shipping and handling. If the total order is 54.34 how much was the order total before shi
Andrews [41]

Answer:

$52

Step-by-step explanation:

Given that :

Amount charged for shipping and handling = 4.5%

Let cost of order = x

Total amount charged = 54.34

However,

total. Amount charged = (total cost + handling and shipping)

54.34 = x + 0.045x

54.34 = 1.045x

Order total before shipping = 54.34 / 1.045

Order total before shipping = 52

6 0
3 years ago
Pls tell me the answer with the work shown
shepuryov [24]

Answer:

plug in the numbers for radius, and know your formulas. a diameter is 2 times a radius so

Pi or 3.14 times D is the same as 2 times pi times radius.

4 0
3 years ago
Brent is a researcher for a food company. He is on a team creating a reduced-calorie version of its flagship cracker. The team w
Andre45 [30]

Answer:

Step-by-step explanation:

Hello!

The research team created a cracker with fewer calories. The average content of calories of the new crackers per serving of 6 should be less than 60.

To test it a random sample of 26 samples of the new cracker was taken and the calories per serving were measured.

Then the study variable is

X: calories of a 6 serve sample of the new reduced-calorie version. (cal)

The variable has a normal distribution with a population standard deviation of 0.82 cal.

To test the claim that the new crackers have on average less than 60 calories, the parameter of interest is the population mean (μ) and the hypotheses are:

H₀: μ ≥ 60

H₁: μ < 60

α: 0.01

Since the variable has a normal distribution and the population variance is known, the best statistic to use to conduct the test is a Standard Normal

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

This test is one tailed to the left, wich means that the null hypothesis will be rejected at low levels of the statistic.

Z_{\alpha } = Z_{0.01} = -2.334

If Z ≤ -2.334, the decision is to reject the null hypothesis.

If Z > -2.334, the decision is to not reject the null hypothesis.

Using the data of the sample I've calculated the sample mean.

X[bar]= ∑X/n= 1548.61/26= 59.56 cal

Z_{H_0}= \frac{(59.56-60)}{\frac{0.82}{\sqrt{26} } } = -2.736

The observed Z value is less than the critical value, so the decision is to reject the null hypothesis.

At a level of significance of 1%, you can conclude that the population mean of calories of the samples of new crackers is less than 60 cal.

I hope it helps!

6 0
3 years ago
Name 3 benefits of trees
alexandr402 [8]
They get carbon dioxide out of the air

they give us oxygen

they are beautiful and provide homes and food for animals

Hope this helped
3 0
3 years ago
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