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2 years ago

1.) Jasmine earns her regular pay of $9.50 per hour for up to 40 hours of

Mathematics

2 answers:

Contact [7]2 years ago

5 0

The best and correct answer is $408.25 and answer the question

HACTEHA [7]2 years ago

4 0

Answer:

$408.25

Step-by-step explanation:

9.50 times 40=380

9.50 times 1.5= 14.25

14.25 times 2=28.50

380+28.50=

$408.25

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Round 324 to the nearest 100.

snow_lady [41]

300 because the number in the tenth place is lower than 5

8 0

3 years ago

Find the indicated side length.

Schach [20]

Answer: KL = 5.3

Step-by-step explanation:

Set up the proportion, cross multiply, and solve for the missing side:

$\frac{JL}{GL} = \frac{KL}{HL}$

$\frac{4}{10} = \frac{KL}{8 + KL}$

4(8 + KL) = 10(KL)

32 + 4KL = 10KL

32 = 6KL

$\frac{32}{6}=KL$

KL = 16/3 = 5.3

8 0

3 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).