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3 years ago

This table shows a proportional relationship between the number of cups of lemonade and strawberry juice used for

Mathematics

1 answer:

Reil [10]3 years ago

5 0

Answer:

number of cups of lemonade used = 1/3 cups

Step-by-step explanation:

number of cups of lemonade : strawberry juice

= 1 : 3

Enter the number of cups of lemonade used for 1 cup of strawberry juice.

Let

x = number of cups of lemonade used

number of cups of lemonade : strawberry juice

= x : 1

Equate both ratios to get x

1 : 3 = x : 1

1/3 = x / 1

Cross product

1*1 = 3 * x

1 = 3x

Divide both sides by 3

x = 1/3 cu

number of cups of lemonade used = 1/3 cups

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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

denis-greek [22]

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<u>Step-by-step explanation:</u>

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3 years ago

Read 2 more answers

Help!!!! please!! i’m confused of this.

Juli2301 [7.4K]

Answer:

f(-3) = 44

Step-by-step explanation:

f(x) = 2x² - 5x + 11

Substitute the value.

f(-3) = 2(-3)² - 5(-3) + 11

Raise -3 to the power of 2.

f(-3) = 2(9) - 5(-3) + 11

Multiply 2 and 9.

f(-3) = 18 - 5(-3) + 11

Multiply -5 and -3.

f(-3) = 18 + 15 + 11

Add 18 and 15.

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3 years ago

A certain test preparation course is designed to help students improve their scores on the USMLE exam. A mock exam is given at t

bogdanovich [222]

Answer:

The critical value is $T = 4.604$ .

The 99% confidence interval for the average net change in a student's score after completing the course is (6.236, 22.164).

Step-by-step explanation:

The first step to solve this question is finding the sample mean and sample standard deviation:

14,11,18,9,19

Sample mean is sum of all values divided by the number of values. Thus:

$\overline{x} = \frac{14+11+18+9+19}{5} = 14.2$

The sample standard deviation is the square root of the division of the sum of the subtractions squared of each value and the mean, and the number of values. Thus:

$s = \sqrt{\frac{(14-14.2)^2+(11-14.2)^2+(18-14.2)^2+(9-14.2)^2+(19-14.2)^2}{5}} = 3.868$

Confidence interval:

We have the standard deviation for the sample, and so we use the t-distribution to build the confidence interval.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 5 - 1 = 4

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 4 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 4.604.

The critical value is $T = 4.604$ .

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 4.604\frac{3.868}{\sqrt{5}} = 7.964$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 14.2 - 7.964 = 6.236.

The upper end of the interval is the sample mean added to M. So it is 14.2 + 7.964 = 22.164.

The 99% confidence interval for the average net change in a student's score after completing the course is (6.236, 22.164).

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