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3 years ago

What’s the slope of this line?

Mathematics

2 answers:

arlik [135]3 years ago

7 0

I think it might be 9/8

madreJ [45]3 years ago

3 0

-8/9i think it might be -8/9

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I need help this question?

Masja [62]

Answer:

Correct answer: F. graph F or x ∈ |-5 ; 5| (including endpoints)

Step-by-step explanation:

Let us first define the absolute value:

| x | = 1. { x with condition x ≥ 0 }

or 2. { - x with condition x < 0 }

This is a linear inequality

1. x ≤ 5 ∧ x ≥ 0 ⇒ 0 ≤ x ≤ 5 or interval x ∈ |0 ; 5| (including endpoints)

2. - x ≤ 5 when we multiply both sides of the equation by -1 we get:

x ≥ -5 ∧ x < 0 ⇒ -5 ≤ x < 0 or interval x ∈ |-5 ; 0) (including -5)

The solution to this linear inequality is the union of these two intervals:

x ∈ |-5 ; 0) ∪ |0 ; 5| ⇒ x ∈ |-5 ; 5| (including endpoints)

x ∈ |-5 ; 5| (including endpoints)

God is with you!!!

7 0

3 years ago

A random sample of 20 students yielded a mean of ¯x = 72 and a variance of s2 = 16 for scores on a college placement test in mat

Helen [10]

Answer:

The 98% confidence interval for the variance in the pounds of impurities would be $8.400 \leq \sigma^2 \leq 39.827$ .

Step-by-step explanation:

1) Data given and notation

$s^2 =16$ represent the sample variance

s=4 represent the sample standard deviation

$\bar x$ represent the sample mean

n=20 the sample size

Confidence=98% or 0.98

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=20-1=19$

Since the Confidence is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and we can use excel, a calculator or a table to find the critical values.