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Korvikt [17]
3 years ago
7

Plz help this is confusing

Mathematics
1 answer:
Ahat [919]3 years ago
6 0

Answer: $ 1.50

Step-by-step explanation:

Sorry if I’m wrong

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9. Write an equation for the table in slope-
never [62]
The slope is going to be 1.6
So y = 1.6x - 5
6 0
3 years ago
PLEASE HELP ME!!!!
chubhunter [2.5K]
We know, Point slope form is written as: 
y - y₁ = m(x - x₁)

First, We need to Calculate the slope:
m = (y₂ - y₁) / (x₂ - x₁)
m = (2 - 1) / (-3-2)
m = 1/-5
m = -1/5

Take any one of them: Let's take (-3, 2) as our coordinate:
y - 2 = m(x - (-3))
y - 2 = -1/5(x + 3)

In short, Your Answer would be Option D

Hope this helps!
4 0
4 years ago
Annual starting salaries in a certain region of the U. S. for college graduates with an engineering major are normally distribut
defon

Answer:

0.8665 = 86.65% probability that the sample mean would be at least $39000

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 normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, 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 normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean $39725 and standard deviation $7320.

This means that \mu = 39725, \sigma = 7320

Sample of 125:

This means that n = 125, s = \frac{7320}{\sqrt{125}} = 654.72

The probability that the sample mean would be at least $39000 is about?

This is 1 subtracted by the pvalue of Z when X = 39000. So

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

By the Central Limit Theorem

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

Z = \frac{39000 - 39725}{654.72}

Z = -1.11

Z = -1.11 has a pvalue of 0.1335

1 - 0.1335 = 0.8665

0.8665 = 86.65% probability that the sample mean would be at least $39000

4 0
3 years ago
I need the answer quick
Vanyuwa [196]

Answer:

the order pairs show us how many feet the hot air balloon rises in a certain amount of seconds

Step-by-step explanation:

6 0
3 years ago
What is the following product? (2 square root 7+3 square root 6)(5 square root 2+4 square root 3)
Xelga [282]

(2\sqrt7+3\sqrt6)(5\sqrt2+4\sqrt3)\qquad\text{use distributive property}\\\\=(2\sqrt7)(5\sqrt2)+(2\sqrt7)(4\sqrt3)+(3\sqrt6)(5\sqrt2)+(3\sqrt6)(4\sqrt3)\\\\=10\sqrt{14}+8\sqrt{21}+15\sqrt{12}+12\sqrt{18}\\\\=10\sqrt{14}+8\sqrt{21}+15\sqrt{4\cdot3}+12\sqrt{9\cdot2}\\\\=10\sqrt{14}+8\sqrt{21}+15\sqrt4\cdot\sqrt3+12\sqrt9\cdot\sqrt2\\\\=10\sqrt{14}+8\sqrt{21}+(15)(2)\sqrt3+(12)(3)\sqrt2\\\\=\boxed{10\sqrt{14}+8\sqrt{21}+30\sqrt3+36\sqrt2}

5 0
3 years ago
Read 2 more answers
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