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AveGali [126]
2 years ago
15

At her restaurant .Yolanda open 285 cans of tomato sauce .She need to use 4 can of for one pan of lasagna how many pans of lasag

na could yolanda make if she uses all if she all the cans what is the answer in a mixed numder
Mathematics
1 answer:
Rudiy272 years ago
8 0

Answer:

71 1/4 pans of lasagna

Step-by-step explanation:

4 cans of tomato sauce = 1 pan of lasagna

285 cans of tomato sauce = x pan of lasagna

4 = 1

285 = x

Cross product

4*x = 285*1

4x = 285

x = 285/4

x = 71 1/4 pans of lasagna

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Arte-miy333 [17]

Answer:

7

Step-by-step explanation:

4 0
3 years ago
Help please, and explain.
Korvikt [17]

Answer: 2x + y

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

logₐ(3) = x

logₐ(5) = y

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             = 2 logₐ(3) + logₐ(5)

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4 0
3 years ago
Read 2 more answers
Find the value of c that makes the trinomial a perfect square trinomial.<br> m^2-14m+c
Pie

Answer:

49

Step-by-step explanation:

m^2-14m+c

Take the -14

Divide by 2

-14/2 = -7

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(-7)^2 = 49

Add it to each side

m^2-14m+49

The value of c is 49

6 0
3 years ago
Someone helped me please
suter [353]

Answer:

Step-by-step explanation:

add all student to get total.

30+28+14+28=100 total students

28+14+28=70 students that do not supprt ashley.

70/100 is the probability the student chosen at random will not support ashley.

.7 in simplified decimal format.

4 0
2 years ago
The weight of baby goats is believed to be Normally distributed, with a mean of 5.75 pounds. The average weight of a random samp
Julli [10]

Answer:

The standard error of the mean is 0.0783.

Step-by-step explanation:

The Central Limit Theorem helps us find the standard error of the mean:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

The standard deviation of the sample is the same as the standard error of the mean. So

SE_{M} = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\sigma = 0.35, n = 20

So

SE_{M} = \frac{\sigma}{\sqrt{n}}

SE_{M} = \frac{0.35}{\sqrt{20}}

SE_{M} = 0.0783

The standard error of the mean is 0.0783.

7 0
3 years ago
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