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dem82 [27]
4 years ago
14

Hi can someone plz help me with #1...it’s being graded. sos thanks so much!!

Mathematics
2 answers:
olga55 [171]4 years ago
7 0

v= c x p

v= 105 x 25

v= 2,625 calories in total.

densk [106]4 years ago
3 0
1:105::25:X
105*25whole divided by 1*X
2625
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How do you simplify 2/3 divided by 1/3
suter [353]

Answer:

2

Step-by-step explanation:

Solve. To do so, change the division sign into a multiplication sign, and flip the second fraction:

(2/3)/(1/3) = (2/3) x (3/1)

Multiply across:

(2 * 3)/(3 * 1) = (6)/(3)

Simplify:

6/3 = 2

2 is your answer.

~

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3 years ago
How much water should be added to 14mL of 15% alcohol solution to reduce the concentration to 7%
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6 0
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What's the length of side b in the figure below? I posted a picture so you can view the actual problem. Please help.
svetoff [14.1K]

we know that

in the right triangle of the figure

Applying the Pythagorean Theorem

13^{2} =b^{2} +7^{2} \\ b^{2} =13^{2}-7^{2}\\ b^{2}=120\\ b=\sqrt{120} \ cm

therefore

the answer is the option

C) b=\sqrt{120} \ centimeters

6 0
4 years ago
Read 2 more answers
Car A's fuel efficiency is 34 miles per gallon of gasoline, and car B's fuel efficiency is 23 miles per gallon of gasoline. At t
Verdich [7]

Answer:

22 more gallons.

Step-by-step explanation:

Given:

Car A's fuel efficiency is 34 miles per gallon of gasoline.

Car B's fuel efficiency is 23 miles per gallon of gasoline.

Question asked:

At those rates, how many more gallons of gasoline would car B consume than car A on a 1,564 miles trip?

Solution:

<u>Car A's fuel efficiency is 34 miles per gallon of gasoline.</u>

<u>By unitary method:</u>

Car A can travel 34 miles in = 1 gallon

Car A can travel 1 mile in = \frac{1}{34} \ gallon

Car A can travel 1564 miles in = \frac{1}{34} \times1564=46\ gallons

<u>Car B's fuel efficiency is 23 miles per gallon of gasoline.</u>

Car B can travel 23 miles in = 1 gallon

Car B can travel 1 mile in = \frac{1}{23} \ gallon

Car B can travel 1564 mile in = \frac{1}{23} \times1564=68\ gallons

We found that for 1564 miles trip Car A consumes 46 gallons of gasoline while  Car B consumes 68 gallons of gasoline that means Car B consumes 68 - 46 = 22 gallons more gasoline than Car A.

Thus, Car B consumes 22 more gallons of gasoline than Car A consumes.

<u />

6 0
3 years ago
The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a st
Alinara [238K]

Answer:

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

Step-by-step explanation:

Problems of normally distributed samples can be 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}

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. The sum of the probabilities is decimal 1. So 1-pvalue is the probability that the value of the measure is larger than X.

In this problem

The line width used for semiconductor manufacturing is assumed to be normally distributed with a mean of 0.5 micrometer and a standard deviation of 0.05 micrometer, so \mu = 0.5, \sigma = 0.05.

What is the probability that a line width is greater than 0.62 micrometer?

That is P(X > 0.62)

So

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

Z = \frac{0.62 - 0.5}{0.05}

Z = 2.4

Z = 2.4 has a pvalue of 0.99180.

This means that P(X \leq 0.62) = 0.99180.

We also have that

P(X \leq 0.62) + P(X > 0.62) = 1

P(X > 0.62) = 1 - 0.99180 = 0.0082

There is a 0.82% probability that a line width is greater than 0.62 micrometer.

3 0
3 years ago
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