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Maksim231197 [3]
3 years ago
11

Without computing, decide whether the value of each expression is much smaller than one, closer to one, or much greater than one



Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
3 0

Answer:

C: closer to one

D: much smaller than one

E: much greater than one

F: much smaller than one

Step-by-step explanation:

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Find the value of n that makes ΔDEF ∼ΔXYZ when DE = 4, EF = 5, XY = 4(n+1), YZ = 7n - 1, and ∠E ≅∠Y. n =
dalvyx [7]

Answer:

3

Step-by-step explanation:

It's given that ΔDEF ∼ΔXYZ . So the corresponding sides of both triangles will be proportional to each other.

=  > \frac{de}{xy}  =    \frac{ef}{yz}  =  \frac{df}{xz}

DE = 4 ; XY = 4(n + 1) ; EF = 5 ; YZ = 7n - 1

Putting all these values gives ,

\frac{4}{4(n + 1)}  =  \frac{5}{7n - 1}

=  >  \frac{1}{n + 1}  =  \frac{5}{7n - 1}

=  > 7n - 1 = 5(n + 1)

=  > 7n - 1 = 5n + 5

=  > 7n - 5n = 5 + 1

=  > 2n = 6

=  > n =  \frac{6}{2}  = 3

3 0
3 years ago
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At McDonalds,a cheeseburger has 200 fewer calories than a large fries.Two cheeseburgers and a large fries have 1100 calories.How
garik1379 [7]
The fries are 500 calories and each burger is 300 calories. 500+300+300=1100
8 0
4 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!

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