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masha68 [24]
3 years ago
13

Myron is cooking meat. The double number line shows that he needs 1 2 12 hours to cook 5 5 kilograms of meat. 0 0 5 5 Meat Meat

(kg) (kg) 0 0 1 2 12 Time Time (hours) (hours) Based on the ratio shown in the double number line, how long does Myron need to cook 3 3 kilograms of meat?

Mathematics
1 answer:
musickatia [10]3 years ago
4 0

Answer: 7 hours 12 minutes

Step-by-step explanation:

Let he needs x hours to make 1 kg of meat.

Therefore, time taken in making 5 kg of meat = 5 × x = 5x hours.

But, According to the question,

He needs 12 hour to cook 5 kg of meat.

Thus, 5 x = 12

⇒ x = \frac{12}{5} ( By dividing both sides by 5 )

⇒ He needs \frac{12}{5} hours to make 1 kg of meat.

⇒ The time taken to cook 3 kg of meat = 3\times \frac{12}{5}=\frac{36}{5}

= 7\frac{1}{5}  hours ⇒ 7 hours 12 minutes

Hence, He needs 7 hours 12 minutes time to cook 3 kg of meat.

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An art history professor assigns letter grades on a test according to the following scheme. A: Top 13%13% of scores B: Scores be
amm1812

Answer:

The numerical limits for a B grade is between 81 and 89.

Step-by-step explanation:

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.

In this problem, we have that:

\mu = 79.7, \sigma = 8.4

B: Scores below the top 13% and above the bottom 56%

Below the top 13%:

Below the 100-13 = 87th percentile. So below the value of X when Z has a pvalue of 0.87. So below X when Z = 1.127. So

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

1.127 = \frac{X - 79.7}{8.4}

X - 79.7 = 8.4*1.127

X = 89

Above the bottom 56:

Above the 56th percentile, so above the value of X when Z has a pvalue of 0.56. So above X when Z = 0.15. So

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

0.15 = \frac{X - 79.7}{8.4}

X - 79.7 = 8.4*0.15

X = 81

The numerical limits for a B grade is between 81 and 89.

3 0
3 years ago
If f(2) = 13 and f '(x) ≥ 2 for 2 ≤ x ≤ 7, how small can f(7) possibly be?​
timofeeve [1]

Answer:

23

Step-by-step explanation:

We are given that

f(2)=13

f'(x)\geq 2

2\leq x\leq 7

We have to find the possible small value of f(7).

We know that

f'(x)=\frac{f(b)-f(a)}{b-a}

Using the formula

f'(x)=\frac{f(7)-f(2)}{7-2}

f'(x)=\frac{f(7)-13}{5}

We have

f'(x)\geq 2

\frac{f(7)-13}{5}\geq 2

f(7)-13\geq 2\times 5

f(7)-13\geq 10

f(7)\geq 10+13

f(7)\geq 23

The small value of f(7) can be 23.

3 0
3 years ago
Find the smallest integer value of X that satisfies the inequality of 6X>7
Nastasia [14]

Answer:

  x = 2

Step-by-step explanation:

The solution set can be found by dividing the inequality by the coefficient of x.

  6x > 7

  x > 7/6

The smallest integer greater than 7/6 is 12/6 = 2.

The smallest integer solution is x = 2.

5 0
3 years ago
What is the mean of 7, 5, 9, 5, and 12
Kryger [21]
The mean is: 7.6
The range is: 7
The median is: 7
The mode is: 5
:)
5 0
3 years ago
Read 2 more answers
A red blinking light blinks every 18 seconds, while a yellow blinking light blinks every 12 seconds. If both lights start at the
Mice21 [21]
Option C, 36, because it is the LCM (Least common multiple)
4 0
4 years ago
Read 2 more answers
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