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suter [353]
3 years ago
8

Find the x- and y-intercept of each function. 1) x + 2 y = -6

Mathematics
2 answers:
Nimfa-mama [501]3 years ago
8 0

Answer:

x-intercept is-6

y-intercept is -3

Step-by-step explanation:

for x-intercept,y=0,x=-6

for y-intercept,x=0,2y=-6,y=-3

zimovet [89]3 years ago
4 0
X intercept (-6,0) y intercept (0,-3)
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During a food drive, a local middle school collected 8,887 canned food items. Each of the 27 classrooms that participated in the
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8,887 cans divided by 27 classrooms is 329.1481 so each class donated about
329 cans
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2 years ago
How do you work out 7 2/5 -5 3/5 by subtracting mixed numbers with borrowing
Marat540 [252]

Answer:

1 4/5

Step-by-step explanation:

37/5-5 3/5

37/5-28/5

9/5

1 4/5

6 0
3 years ago
The mean weight of an adult is 69 kilograms with a variance of 121. If 31 adults are randomly selected, what is the probability
amid [387]

Answer:

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

Step-by-step explanation:

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

Also, important to remember that the standard deviation is the square root of the variance.

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 = 69, \sigma = \sqrt{121} = 11, n = 31, s = \frac{11}{\sqrt{31}} = 1.97565

What is the probability that the sample mean would be greater than 70.5 kilograms?

This is 1 subtracted by the pvalue of Z when X = 70.5. So

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

By the Central limit theorem

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

Z = \frac{70.5 - 69}{1.97565}

Z = 0.76

Z = 0.76 has a pvalue of 0.7764

1 - 0.7764 = 0.2236

0.2236 = 22.36% probability that the sample mean would be greater than 70.5 kilograms.

8 0
3 years ago
∠1 and ∠2 are complementary angles. m∠1 is 5y+32 and m∠2 is 7y-14. Find m∠2 & show your work.
Ainat [17]
  • m<1+m<2=90

\\ \sf\longmapsto 5y+32+7y-14=90

\\ \sf\longmapsto 12y+18=90

\\ \sf\longmapsto 12y=72

\\ \sf\longmapsto y=6

  • m<2=7(6)-14=42-14=28[/tex]
7 0
3 years ago
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. Hannah is selling slices of pie at the bake sale. The pie has 8 slices. She has sold 1/4 of the slices. What fraction with a d
trasher [3.6K]

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2/8 because each 1/4 0f the 8 slices is equal to 2/8 by multipliction or you can say 1/4 of 8

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2 years ago
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