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3 years ago

Using point-slope form, write the equations of the lines that have the following slopes and points.

Mathematics

1 answer:

zavuch27 [327]3 years ago

8 0

Point slope form: y - y1 = m(x - x1)
Plug in the values
a) y + 4 = 5(x - 2)
b) y - 9 = -3/2(x + 2)

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Using the given set of numbers find the mean (rounded to the nearest tenth), median,mode and range: 18,19,16,18,14,17,18 and 15

sashaice [31]

Answer:

Mean = 16.9
Mode = 18
Median = 17.5

Step-by-step explanation:

Given data:

18, 19, 16, 18, 14, 17, 18, 15

The data in the ascending order:

14, 15, 16, 17, 18, 18, 18, 19

The mean:

(14 + 15 + 16 + 17 + 18*3 + 19)/8 = 16.9 (rounded)

The median:

(17 + 18)/2 = 17.5 (the average of middle terms)

The mode:

18 (the most repeated number)

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3 years ago

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An equation that contains only the operation of addition is called

Llana [10]

Answer: an "addition equation" .
__________________________________________

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3 years ago

Twice the difference of a number and six is less than the number decreased by two

Semenov [28]

What’s the question ??

7 0

3 years ago

The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.

inessss [21]

The lengths of pregnancies are normally distributed with a mean of 266 days and a standard deviation of 15 days.

That is,

Consider X be the length of the pregnancy

Mean and standard deviation of the length of the pregnancy.

Mean $\mu =266\\$

Standard deviation \sigma =15

For part (a) , to find the probability of a pregnancy lasting 308 days or longer:

That is, to find $P(X\geq 308)$

Using normal distribution,

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

$z=\frac{308-266}{15}$

$=\frac{42}{15}$

Thus $z=2.8$

So $P\left (X\geq 308 \right )=1-P(X$

$=1-P(z$

$=1-Table\: value\: of\: 2.8$

$=1-0.99744$

$=0.00256$

Thus the probability of a pregnancy lasting 308 days or longer is given by 0.00256.

This the answer for part(a): 0.00256

For part(b), to find the length that separates premature babies from those who are not premature.

Given that the length of pregnancy is in the lowest 3%.

The z-value for the lowest of 3% is -1.8808

Then $X=\frac{X-\mu}{\sigma}\Rightarrow X=z*\sigma+\mu$

This implies $X=-1.8808*15+266=237.788$

Thus the babies who are born on or before 238 days are considered to be premature.

5 0

3 years ago

What is the place value of 6 in the number 6,029?

Feliz [49]

Answer: 6 on the extreme left has the place value 6000; the next 6 has the value 600; the next, 60; and the last, 6. The numeral for every whole number stands for a sum. 364 = 3 Hundreds + 6 Tens + 4 Ones. Hope this helps!

Step-by-step explanation:

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3 years ago

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