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3 years ago

PLEASEEE HELP ASAP! grade 9 work and show your work!!

Mathematics

1 answer:

lidiya [134]3 years ago

5 0

Answer:

at 0 the amount is 5

1 is 7.5

2 is 10

3 is 12.5

the initial value is the y intercept when the value ox x=0

the initial value is 5

the rate of change is the total change of a function of the out put (y) divided by the change of the input (x)

f(x2)-f(x1)/x2-x1

10-7.5/2-1

2.5/1=2.5

or m=y2-y1/x2/x1=10-7.5/2-1=2.5

y=2.5x+5

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Find the percent incease or decrease: From 32 points to 46 points There was a increase

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Answer:

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3 years ago

Is 19,279 divisibility by 2

erastovalidia [21]

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3 years ago

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2(x-1) ≥ 10 or 3-4x > 15<br> Solve the compound inequality, and show work!

Veseljchak [2.6K]

Answer:

$x\geq 6\text{ or } x$

Step-by-step explanation:

We have the compound inequality:

$2(x-1)\geq10\text{ or } 3-4x>15$

Let's solve each of them individually first:

We have:

$2(x-1)\geq10$

Divide both sides by 2:

$x-1\geq5$

Add 1 to both sides:

$x\geq6$

We have:

$3-4x>15$

Subtract from both sides:

$-4x>12$

Divide both sides by -4:

$x$

Hence, our solution set is:

$x\geq 6\text{ or } x$

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3 years ago

Read 2 more answers

A random sample of 100 teletype operators indicates that their salaries fluctuate quite a bit. The sample Standard deviation of

ELEN [110]

Answer:

$\frac{(99)(10)^2}{123.225} \leq \sigma^2 \leq \frac{(99)(10)^2}{77.046}$

$80.341 \leq \sigma^2 \leq 128.495$

Now we just take square root on both sides of the interval and we got:

$8.963 \leq \sigma \leq 11.336$

Step-by-step explanation:

Data given and notation

s=10 represent the sample standard deviation

$\bar x$ represent the sample mean

n=100 the sample size

Confidence=90% or 0.90

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 .

Calculating the confidence interval

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

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

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

$df=n-1=100-1=99$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,99)" "=CHISQ.INV(0.95,99)". so for this case the critical values are:

$\chi^2_{\alpha/2}=123.225$

$\chi^2_{1- \alpha/2}=77.046$

And replacing into the formula for the interval we got:

$\frac{(99)(10)^2}{123.225} \leq \sigma^2 \leq \frac{(99)(10)^2}{77.046}$

$80.341 \leq \sigma^2 \leq 128.495$

Now we just take square root on both sides of the interval and we got:

$8.963 \leq \sigma \leq 11.336$

5 0

3 years ago

Please help!!! This really affects my grade! I need a final answer not a guess. Thank you very much. I really appreciate it. Lin

Darina [25.2K]

Answer:

Left skewed

Step-by-step explanation:

Looking at the box and whisker plot attached showing the wait times and the number of times they occur, we can see that as we approach the left, the number of times the wait time occurs increases. The box plot also shows like as if the box was pushed to the right as the center line is closer to the right than left.

Thus, we can say it's left skewed.

8 0

3 years ago

PLEASEEE HELP ASAP! *grade 9 work* and show your work!!

PLEASEEE HELP ASAP! grade 9 work and show your work!!