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3 years ago

Please help asap 30 pts

Mathematics

2 answers:

aliya0001 [1]3 years ago

8 0

$5y+1.8=4y-3.2\qquad|\text{subtract 1.8 from both sides}\\\\5y=4y-5\qquad|\text{subtract 4y from both sides}\\\\\boxed{y=-5}\to\boxed{a.}$

Semmy [17]3 years ago

4 0

Step 1. Subtract 4y from both sides

5y + 18 - 4y = -3.2

Step 2. Simplify 5y + 1.8 -4y to y + 1.8

y + 1.8 = -3.2

Step 3. Subtract 1.8 from both sides

y = -3.2 - 1.8

Step 4. Simplify -3.2 - 1.8 to -5

y = -5

Your answer is A.

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The following data were collected from 12 rain gauges in a park. Build a 95% CI for the mean rainfall at the park.

dybincka [34]

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Critical values: $t_{\alpha/2}=-2.201$ $t_{1-\alpha/2}=2.201$

95% confidence interval would be given by (3.646;4.472)

Step-by-step explanation:

1) Previous concepts

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 means 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.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The data is:

4.65 3.89 2.73 4.35 3.80 4.86 4.33 4.37 4.76 4.05 3.05 3.87

2) Compute the sample mean and sample standard deviation.

In order to calculate the mean and the sample deviation we need to have on mind the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$

=AVERAGE(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

On this case the average is $\bar X= 4.059$

=STDEV.S(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)

The sample standard deviation obtained was s=0.6503

3) Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =12 <30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . The degrees of freedom are given by: