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3 years ago

1. Instant Meals sent out free samples to introduce its new product, Sesame soup. Each

1 answer:

5 0

Hello there! For this word problem we first have to identify the numbers and variables if there are any. Our numbers are 36 cents per 23 ounces, each sample weighs 69 ouches, and how much it costs to mail out one hundred eighty-eight.

Next, let's see how many 23 ounces are in 69 ounces. To solve for that, we can divide 69 by 23. 63 divided by 23 is 3, so there is 3 sets of 23 ounces per 69 ounce samples. Then, we can multiply 3 by 36 cents in order to see how much each sample costs. 3 times .36 is 1.08 or 1 dollar and 8 cents. Lastly, we need to multiply 188 by 1.08 to figure out how much it costs to mail out 188 samples.

188 times 1.08 equals to 203.04 or 203 dollars and 4 cents. Therefore, it would cost $203.04 to mail out 188 samples. Hope this helps! Have a great day!

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mina [271]

Answer:

Step-by-step explanation:

A. 2 meters to centimeters

1 m = 100 cm

$2 m = 2 \times 100 \\2 m = 200 cm = 2 \times 10^2$

B)108 centimeters to meters

100 cm =1 m

$1 cm = \frac{1}{100} m \\108 cm = \frac{108}{100}=108 \times 10^{-2} m$

C)2.49 meters to centimeters

1 m = 100 cm

$2.49 m = 2.49 \times 100 = \frac{249}{100} \times 100 = 249 cm$

D)50 centimeters to meters

100 cm = 1m

$1 cm = \frac{1}{100} m \\50 cm = \frac{50}{100}= 5 \times 10^{-1} m$

E)6.3 meters to centimeters

1 m = 100 cm

$6.3 m = 6.3 \times 100=\frac{63}{10} \times 100 = 630 cm$

F)7 centimeters to meters

100 cm = 1 m

$1 cm = \frac{1}{100} m\\\\7 cm = \frac{7}{100} = 7 \times 10^{-2} m$

a)4 meters to millimeters

1 m = 1000 mm

$4 m = 4000 = 4 \times 10^3 mm$

b)1.7 meters to millimeters

1 m = 1000 mm

$1.7 m = 1.7 \times 1000= \frac{17}{10} \times 1000=17 \times 10^2 mm$

3 0

3 years ago

Help me with this pls

vladimir2022 [97]

90 counterclockwise hope I’m right

8 0

2 years ago

Read 2 more answers

Ms hills travels 2400 miles at a rate of 400 miles per day to visit her grandchild . Write and solve an equation to find the num

Naya [18.7K]

It Would Take her 6 days

2400÷400=6

6 0

3 years ago

8x(5+10)= (8x5) + (5x10)

Slav-nsk [51]

Answer:

i think its x= 0.75

Step-by-step explanation:

substitute:

8x × 5 = 40x

8x × 10 = 80x

8 × 5 = 40

5 × 10 =50

40x + 80x = 40 + 50

add like terms

so it'd be:

120x = 90

now divide 90 by 120 and that would give you the value of x alone which is 0.75

7 0

3 years ago

Read 2 more answers

The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand

Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 18.6, \sigma = 5.9$

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{21 - 18.6}{5.4}$

$Z = 0.44$

$Z = 0.44$ has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have $n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768$