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LenKa [72]
3 years ago
15

What is the x-intercept of the line with the equation y=1/2x-3

Mathematics
1 answer:
EleoNora [17]3 years ago
7 0
A line crosses the x axis when y=0. the value of x at that point is the x-intercept

so substitute 0 for y and solve for x

0=1/2x-3

i like to multiply everything by the denominator of a fraction to clear fractions out

2(0)=2(1/2)x-2(3)

0=x-6
6=x
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Choosy Chocolates sold chewy squares and chocolate stars at a ratio of 3 to 8. 45 more chocolate stars were sold than chewy squa
masya89 [10]
So 45 more chocolate stars were sold compared to chewy squares, which makes it a ratio of 3:8, hence we can assume that (45÷5=9) is the amount sold for each ratio.

If you multiply 9 by (3+8=11), you will get 99.
3 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
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18 has more factors than all of those numbers.

1, 2, 3, 6, 9, 18 are all the factors of 18.

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Compare −114 and −1 greater than or less than
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Answer:

-114 is less than -1

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2/3 = 8/12
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You start with 8/12 and eat 4/12 so 4/12 (or one quarter) remains.
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