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sasho [114]
3 years ago
13

Hellpppppppppppppppppppppp

Mathematics
2 answers:
xxTIMURxx [149]3 years ago
4 0

Answer:

1. MIGHT be 16 1/2

Step-by-step explanation:

causse 2/3 / 4/9 = 1 1/2

1 1/2 + 10= 11 +5 = 16

16+1/2 = 16 1/2

tatuchka [14]3 years ago
4 0

Answer:

9. 16\frac{1}{9}

Step-by-step explanation:

1. For number 9, start by making 10\frac{2}{3} into a fraction, which would be \frac{32}{3}. Then, find a common denominator. Multiply both sides by 3 to get a common denominator. You should get \frac{96}{9}.

2. Add \frac{96}{9}+\frac{4}{9} to get \frac{100}{9}.

3. Add 5:

5=\frac{45}{9} so our equation is \frac{100}{9}+ \frac{45}{9}, which is \frac{145}{9}.

4. Simplify: 16\frac{1}{9}

----------------------------------------------------------------------------------------------------------------

5. For number 11, start by translating the mixed numbers into fractions:

\frac{9}{2} +\frac{22}{3} * \frac{8}{5}

6. Use PEMDAS to solve:

Multiply- \frac{22}{3} *\frac{8}{5} = \frac{176}{15}

Add- \frac{176}{15} +\frac{9}{2}

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9. (-2, 8), (-6,0)
blsea [12.9K]

Answer: (-2,8); (-6,0)

Step-by-step explanation:

y=mx+b

m=slope

slope=y2-y1/x2-x1

so m=2

y=2x+b

3 0
4 years ago
What is the greatest common faactor of 72 and 40
rusak2 [61]
8×9=72
5×8=40
So the answer is 8
4 0
3 years ago
Read 2 more answers
What price do farmers get for the peach crops? in the third week of June, a random sample of 40 farming regions gave a sample me
prohojiy [21]
Given:
Sample size, n = 40
Sample mean, xb = $6.88
Population std. deviation, σ = $1.92 (known)
Confidence interval = 90%

Assume normal distribution for the population.
The confidence interval is
(xb + 1.645*(σ/√n), xb - 1.645*(σ/√n)
= (6.88 + (1.645*1.92)/√40, 6.88 - (1.645*1.92)/√40)
= (7.38, 6.38)

Answer: The 90% confidence interval is (7.38, 6.38)
6 0
3 years ago
Jerry's team has played 8 games of basketball. During those 8 games his team scored 59, 66, 56, 63, 42, 55, 62 and 51. What is t
zavuch27 [327]

Answer:

To calculate the mean of a set of data, you have to work out the sum of the data (in this case, you need to work out the sum of all the points) and divide it by the number of data points there are (in this case, number of games played).

So you have to do:

(67 + 45 + 84 + 55 + 73 + 36 + 80 + 62 + 38)/9 = 60

The mean number of points is 60.

Step-by-step explanation:

please mark as brainiest

8 0
4 years ago
Read 2 more answers
The weights of college football players are normally distributed with a mean of 200 pounds and a standard deviation of 50 pounds
Feliz [49]

Answer:

0.3811 = 38.11% probability that he weighs between 170 and 220 pounds.

Step-by-step explanation:

When the distribution is normal, we use 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 question, we have that:

\mu = 200, \sigma = 50

Find the probability that he weighs between 170 and 220 pounds.

This is the pvalue of Z when X = 220 subtracted by the pvalue of Z when X = 170.

X = 220

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

Z = \frac{220 - 200}{50}

Z = 0.4

Z = 0.4 has a pvalue of 0.6554

X = 170

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

Z = \frac{170 - 200}{50}

Z = -0.6

Z = -0.6 has a pvalue of 0.2743

0.6554 - 0.2743 = 0.3811

0.3811 = 38.11% probability that he weighs between 170 and 220 pounds.

6 0
4 years ago
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