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What is the answer to this equation 50x^2-32y^2/-5x-4y

LenKa [72]

-10x +8y
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-5x - 4y ) 50x^2 - 32y^2
50x^2 + 40xy
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-40xy - 32y^2
-40xy -32y^2
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answer is -10x + 8y

5 0

3 years ago

Read 2 more answers

Which of the following is the correct graph of the solution to the inequality -13> -5x+2> -28?

anyanavicka [17]

Answer:

first option

Step-by-step explanation:

solving the inequality

- 13 > - 5x + 2 > - 28 ( subtract 2 from each interval )

- 15 > - 5x > - 30

divide each interval by - 5 , reversing the symbols as a result of dividing by a negative quantity.

3 < x < 6

since the symbol is < then end values will be an open circle with the values between shaded blue, that is the first graph

4 0

2 years ago

1. A food safety guideline is that the mercury in fish should be below 1 part per million (ppm). Listed below are the amounts o

Sauron [17]

Answer:

The confidence interval estimate of the population mean is :

(0.61 ppm, 0.90 ppm)

The correct option is (A).

Step-by-step explanation:

The amounts of mercury (ppm) found in tuna sushi sampled at different stores in a major city are:

S = {0.58, 0.82, 0.10, 0.98, 1.27, 0.56, 0.96}

A $(100-\alpha )\%$ confidence interval for the population mean (μ) is an interval estimate of the true value of the mean. This interval has a $(100-\alpha )\%$ probability of consisting the true value of mean.

⇒ Since the population standard deviation is not provided we will use the t-distribution to construct the 99% confidence interval for mean.

⇒ The formula for confidence interval for the population mean is:

$\bar x\pm t_{\alpha/2,(n-1)}\times \frac{s}{\sqrt{n}}$

Here,

$\bar x$ = sample mean

s = sample standard deviation

n = sample size

$t_{\alpha/2,(n-1)}$ = critical value.

The degrees of freedom for the critical value is, (n - 1) = 7 - 1 = 6.

The significance level is: $\alpha =1-Confidence\ level=1-0.99=0.01$

The critical value is:

$t_{\alpha/2,(n-1)}=t_{0.01/2, 7}=t_{0.05,7}=3.143$

**Use the t-table for the critical value.

Compute the sample mean and sample standard deviation as follows:

$\bar x=\frac{1}{7}(0.58+ 0.82+ 0.10+ 0.98+ 1.27+ 0.56+ 0.96) =0.753$

$\int\limits^a_b {x} \, dx s=\sqrt{\frac{\sum (x{i}-\bar x)^{2}}{n-1} } =\sqrt{\frac{1}{6} \times 0.859743} =0.379$

The 99% confidence interval for μ is:

$x^{2} CI=0.753\pm 3.143\times\frac{0.379}{\sqrt{7}} \\=0.753\pm0.143\\=(0.61, 0.896)\\\approx(0.61, 0.90)$

The confidence interval estimate of the population mean is:

(0.61 ppm, 0.90 ppm)

The upper and lower limit of the 99% confidence interval indicates that the true mean value is less than 1 ppm. This implies that there is not too much mercury in tuna sushi

Because it is possible that the mean is not greater than 1 ppm. Also, at least one of the sample values is less than 1 ppm, so at least some of the fish are safe.

Thus, the correct option is (A).