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2 years ago

Natalie is solving a problem. What are some steps she might take after reaching her solution?

1 answer:

8 0

The answer is options C and D

When you are checking over your work and solution, you C) want to make sure you include the correct units so that the answer makes sense in the context of the question and D) see if the answer is reasonable. For example, if I say 2(3)+1=6839, does this make sense? No, especially because the numbers in the question are small integers and could not result in such a large value.

-E :)

You might be interested in

Two random samples are taken from private and public universities

kati45 [8]

Answer:

Step-by-step explanation:

For private Institutions,

n = 20

Mean, x1 = (43120 + 28190 + 34490 + 20893 + 42984 + 34750 + 44897 + 32198 + 18432 + 33981 + 29498 + 31980 + 22764 + 54190 + 37756 + 30129 + 33980 + 47909 + 32200 + 38120)/20 = 34623.05

Standard deviation = √(summation(x - mean)²/n

Summation(x - mean)² = (43120 - 34623.05)^2+ (28190 - 34623.05)^2 + (34490 - 34623.05)^2 + (20893 - 34623.05)^2 + (42984 - 34623.05)^2 + (34750 - 34623.05)^2 + (44897 - 34623.05)^2 + (32198 - 34623.05)^2 + (18432 - 34623.05)^2 + (33981 - 34623.05)^2 + (29498 - 34623.05)^2 + (31980 - 34623.05)^2 + (22764 - 34623.05)^2 + (54190 - 34623.05)^2 + (37756 - 34623.05)^2 + (30129 - 34623.05)^2 + (33980 - 34623.05)^2 + (47909 - 34623.05)^2 + (32200 - 34623.05)^2 + (38120 - 34623.05)^2 = 1527829234.95

Standard deviation = √(1527829234.95/20

s1 = 8740.22

For public Institutions,

n = 20

Mean, x2 = (25469 + 19450 + 18347 + 28560 + 32592 + 21871 + 24120 + 27450 + 29100 + 21870 + 22650 + 29143 + 25379 + 23450 + 23871 + 28745 + 30120 + 21190 + 21540 + 26346)/20 = 25063.15

Summation(x - mean)² = (25469 - 25063.15)^2+ (19450 - 25063.15)^2 + (18347 - 25063.15)^2 + (28560 - 25063.15)^2 + (32592 - 25063.15)^2 + (21871 - 25063.15)^2 + (24120 - 25063.15)^2 + (27450 - 25063.15)^2 + (29100 - 25063.15)^2 + (21870 - 25063.15)^2 + (22650 - 25063.15)^2 + (29143 - 25063.15)^2 + (25379 - 25063.15)^2 + (23450 - 25063.15)^2 + (23871 - 25063.15)^2 + (28745 - 25063.15)^2 + (30120 - 25063.15)^2 + (21190 - 25063.15)^2 + (21540 - 25063.15)^2 + (26346 - 25063.15)^2 = 1527829234.95

Standard deviation = √(283738188.55/20

s2 = 3766.55

This is a test of 2 independent groups. Let μ1 be the mean out-of-state tuition for private institutions and μ2 be the mean out-of-state tuition for public institutions.

The random variable is μ1 - μ2 = difference in the mean out-of-state tuition for private institutions and the mean out-of-state tuition for public institutions.

We would set up the hypothesis. The correct option is

-B. H0: μ1 = μ2 ; H1: μ1 > μ2

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(x1 - x2)/√(s1²/n1 + s2²/n2)

t = (34623.05 - 25063.15)/√(8740.22²/20 + 3766.55²/20)

t = 9559.9/2128.12528473889

t = 4.49

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [8740.22²/20 + 3766.55²/20]²/[(1/20 - 1)(8740.22²/20)² + (1/20 - 1)(3766.55²/20)²] = 20511091253953.727/794331719568.7114

df = 26

We would determine the probability value from the t test calculator. It becomes

p value = 0.000065

Since alpha, 0.01 > than the p value, 0.000065, then we would reject the null hypothesis. Therefore, at 1% significance level, the mean out-of-state tuition for private institutions is statistically significantly higher than public institutions.

4 0

1 year ago

Alaina’s sugar cookie recipe calls for 21 4 cups of flour per batch. If she wants to make 2 3 a batch of cookies, how much flour

Over [174]

"Alaina’s sugar cookie recipe calls for 2 1/4

cups of flour per batch. If she wants to make 2/3

a batch of cookies, how much flour should she use?"

1 1/2 Cups, if she wants to make less than the original recipe, she would need less flour, you have to divide.

8 0

1 year ago

Read 2 more answers

The population of lengths of aluminum-coated steel sheets is normally distributed with a mean of 30.05 inches and a standard dev

Nataliya [291]

Answer:

(a) Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = 0.15716

(b) Probability that a standard normal random variable will be between .3 and 3.2 = 0.3814

Step-by-step explanation:

We are given that the population of lengths of aluminum-coated steel sheets is normally distributed with;

Mean, $\mu$ = 30.05 inches and Standard deviation, $\sigma$ = 0.2 inches

Let X = A sheet selected at random from the population

Here, the standard normal formula is ;

Z = $\frac{X - \mu}{\sigma}$ ~ N(0,1)

(a) The Probability that a sheet selected at random from the population is between 30.25 and 30.65 inches long = P(30.25 < X < 30.65)

P(30.25 < X < 30.65) = P(X < 30.65) - P(X <= 30.25)

P(X < 30.65) = P( $\frac{X - \mu}{\sigma}$ < $\frac{30.65 - 30.05}{0.2}$ ) = P(Z < 3) = 1 - P(Z >= 3) = 1 - 0.001425

= 0.9985

P(X <= 30.25) = P( $\frac{X - \mu}{\sigma}$ <= $\frac{30.25 - 30.05}{0.2}$ ) = P(Z <= 1) = 0.84134

Therefore, P(30.25 < X < 30.65) = 0.9985 - 0.84134 = 0.15716 .

(b) Let Y = Standard Normal Variable is given by N(0,1)

Which means mean of Y = 0 and standard deviation of Y = 1

So, Probability that a standard normal random variable will be between 0.3 and 3.2 = P(0.3 < Y < 3.2) = P(Y < 3.2) - P(Y <= 0.3)

P(Y < 3.2) = P( $\frac{Y - \mu}{\sigma}$ < $\frac{3.2 - 0}{1}$ ) = P(Z < 3.2) = 1 - P(Z >= 3.2) = 1 - 0.000688

= 0.99931

P(Y <= 0.3) = P( $\frac{Y - \mu}{\sigma}$ <= $\frac{0.3 - 0}{1}$ ) = P(Z <= 0.3) = 0.61791

Therefore, P(0.3 < Y < 3.2) = 0.99931 - 0.61791 = 0.3814 .

3 0

1 year ago

A large city in the ancient world was square-shaped. measured in miles, its area numerically exceeded its perimeter by about 132

melisa1 [442]

Area = perimeter + 132.

Let each side of the city be x miles long, then:-

x^2 = 4x + 132
x^2 - 4x - 132 = 0

x = [-(-4) +/- sqrt((-4)^2 - 4 * 1 *-132)] / 2

x = 13.66, -9.66 We ignore the negative

So the city has dimension of 13.66 * 13.66

13.7 * 13.7 to nearest 10th

7 0

1 year ago

Hailey paid \$13$13dollar sign, 13 for 1\dfrac3{7} \text{ kg}1 7 3 kg1, start fraction, 3, divided by, 7, end fraction, start

hichkok12 [17]

Answer:

1kg of salami cost $9.1

Step-by-step explanation:

Hailey paid $13 for 1 3/7 kg of sliced salami.

What was the cost per kilogram of salami?

Cost of 1 3/7 kg of sliced salami=$13

1 3/7 kg=10/7kg

Let x=1 kg of sliced salami

10/7 kg of x=$13

$13=10/7x

13=10/7*x

x=13 ÷ 10/7

=13×7/10

=91/10

=9.1

x=$9.1

Therefore, 1kg of salami cost $9.1

8 0

2 years ago