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2 years ago

Which term is equivalent to this expression

Mathematics

2 answers:

Mariulka [41]2 years ago

8 0

Answer:

Step-by-step explanation:

mel-nik [20]2 years ago

4 0

The answer is letter C

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The pet store had 6 puppies selling for $104 each and 12 kittens selling for $24 each. Today on 2 puppies and 8 kittens were lef

Lunna [17]

You multiply 2*104 that is for puppies and then you do 8 by 24 cause that is for kittens and you get 208:192

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2 years ago

Cheri took out a loan for $14,350. The loan term was 3 years and Cheri’s payments were $625.77 per month. Since Cheri had filed

melamori03 [73]

Answer:

You are correct.

Step-by-step explanation:

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2 years ago

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A bag contains 100 marbles which are red, green, and blue. Suppose a student randomly selects a marble without looking, records

ra1l [238]

Answer:

35 red marbles

Step-by-step explanation:

So we can put the different color marbles in a ratio. So 7:2:11. If we add all of those up we get 20. We do 100/20 to get 5. Each number in the ratio is equivalent to 5 marbles. Then we do 7*5 to get 35. We can check our work by doing, (7*5) + (2*5) + (11*5) = 100. And it does equal 100 so the answer is correct.

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2 years ago

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A random sample of 35 undergraduate students who completed two years of college were asked to take a basic mathematics test. The

lisov135 [29]

Answer:

$(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96$

$(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96$

And the confidence interval would be given by:

$-1.96 \leq \mu_1 -\mu_2 \leq 7.96$

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

Step-by-step explanation:

For this case we have the following info given :

$\bar X_1= 75.1$ represent the sample mean for the scores of the undergraduate students

$s_1 = 12.8$ represent the standard deviation for the undergraduate students

$n_1 =35$ the sample size for the undergraduate

$\bar X_2= 72.1$ represent the sample mean for the scores of the high school students

$s_2 = 14.6$ represent the standard deviation for the high school students

$n_2 =50$ the sample size for the high school

The confidence interval for the true difference of means is given by:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom are given by:

$df=n_1 +n_2 -2= 35+50-2=83$

The confidence level is 90% and the significance level is $\alpha=0.1$ and $\alpha/2 =0.05$ then the critical value would be:

$t_{\alpha/2}= 1.99$

And replacing the info we got:

$(75.1 -72.1) -1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= -1.96$

$(75.1 -72.1) +1.66 \sqrt{\frac{12.8^2}{35} +\frac{14.6^2}{50}}= 7.96$

And the confidence interval would be given by:

$-1.96 \leq \mu_1 -\mu_2 \leq 7.96$

And since the confidence interval contains the value 0 we have enough evidence to conclude that we don't have significant differences between the two means at 10% of significance.

8 0

2 years ago

Garth estimated the height of the door to his classroom in meters what is a reasonable estimate

e-lub [12.9K]

2/3 meters would be a reasonable estimate.

8 0

2 years ago