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2 years ago

14

Lily has $45 in her school lunch account. If she spends $2.50 each day, which graph represents Lily's account balance, y, after

she buys lunch for x days?

2 answers:

n200080 [17]2 years ago

8 0

Y = -2.5x + 45

That's your equation. Your y-intercept would be 45 because that's what she started with. It would be a negative slope as well.

SVEN [57.7K]2 years ago

7 0

Y <span>= -2.50x
(it is negative cuz she is takig out 2.50 each day so her balance is decreasing)

45 </span><span>= -2.50x
</span><span>
so she could by 18 lunches

so since there are no graphs, its the graph with (18,0) as a valid point and a slope of -2.5. hope this makes sense and helps</span>

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The cost of 5 gallons of ice cream has a standard deviation of 8 dollars with a mean of 29 dollars during the summer. What is th

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Answer:

97.74% probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected

Step-by-step explanation:

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

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:

$\mu = 29, \sigma = 8, n = 92, s = \frac{8}{\sqrt{92}} = 0.8341$

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

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

$Z = \frac{27.1 - 29}{0.8341}$

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$Z = -2.28$ has a pvalue of 0.0113

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Answer:

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b) 0.5141 = 51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

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Step-by-step explanation:

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Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

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

$Z = \frac{84.5 - 81.725}{6.6}$

$Z = 0.42$

$Z = 0.42$ has a pvalue of 0.6628.

66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance.

(b) What is the probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance?

Using continuity correction, this is $P(80 - 0.5 \leq X < 90 - 0.5) = P(79.5 \leq X \leq 89.5)$ , which is the pvalue of Z when X = 89.5 subtracted by the pvalue of Z when X = 79.5. So

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(c) What is the probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance?

6.3% of visitors entered through the Grand Lake park entrance, which means that $p = 0.063$

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(d) What is the probability that more than 55 visitors have no recorded point of entry?

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$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.227*0.773} = 5.54$

Using continuity correction, this probability is $P(X \leq 55 + 0.5) = P(X \leq 55.5)$ , which is the pvalue of Z when X = 55.5. So

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

$Z = \frac{55.5 - 39.725}{5.54}$

$Z = 2.85$

$Z = 2.85$ has a pvalue of 0.9978

0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

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koban [17]

Answer:

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Step-by-step explanation:

Since C internally divides AB in the ratio AC/CB = 1/2 = m/n where m = 1 and n = 2, we use the formula for internal division.

Let A = (x₁, y₁) = (5, 16), B = (x₂, y₂) and C = (x, y) = (3, 10)

So x = (mx₂ + nx₁)/(m + n)

y = (my₂ + ny₁)/(m + n)

Substituting the values of the coordinates, we have

x = (mx₂ + nx₁)/(m + n)

3 = (1 × x₂ + 2 × 5)/(2 + 1)

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multiplying through by 3, we have

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2 years ago