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s344n2d4d5 [400]
3 years ago
5

The physician tells a woman who usually drinks 5 cups of coffee daily to cut down on her coffee consumption by 75%. If this woma

n is
compliant with the physician's instruction, how many ounces of coffee is she allowed daily?
Mathematics
1 answer:
Nataly [62]3 years ago
4 0

1 cup=8 ounces

5*8=40

Each day the women drinks 40 oz of coffee.

40oz-75%=10oz

Thus, if the women complies to the physician's instructions, she can drink 10oz or about 1.25 cups of coffee daily.

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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 <em>t</em>-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

<em>s </em>= sample standard deviation

<em>n</em> = sample size

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

The degrees of freedom for the critical value is, (<em>n</em> - 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 <em>t</em>-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).

6 0
3 years ago
Lisa, an experienced shipping clerk, can fill a certain order in 13 hours. Felipe, a
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14 hours or 2 hours. Im not sure
7 0
3 years ago
Write the following equations in slope-intercept form.5x+y=30
Maurinko [17]

Answer:

y=-5x+30

Step-by-step explanation:

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4 0
3 years ago
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Equation
konstantin123 [22]

Answer:

10x+8=4x+(-4).

10×+8=4x-4.

collect like terms.

10x-4x=-4-8.

6x=-12.

x=-12/6.

x=-2.

7 0
3 years ago
Find a particular solution to the nonhomogeneous differential equation y??+4y?+5y=?10x+e^(?x).
Firdavs [7]

Answer:

A) Particular solution:

2x+\frac{1}{2}e^{-x}-\frac{8}{5}

B) Homogeneous solution:

y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))

C) The most general solution is

y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}

Step-by-step explanation:

Given non homogeneous ODE is

y''+4y'+5y=10x+e^{-x}---(1)

To find homogeneous solution:

D^{2}+4D+5=0\\D^{2}+4D+4-4+5=0\\\\(D+2)^{2}=-1\\D+2=\pm iD=-2 \pm i\\y_{h}=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))---(2)

To find particular solution:

y_{p}=Ax+B+Ce^{-x}\\\\y'_{p}=A-Ce^{-x}\\y''_{p}=Ce^{-x}\\

Substituting y_{p},y'_{p},y''_{p} in (1)

y''_{p}+4y'_{p}+5y_{p}=10x+e^{-x}\\Ce^{-x}+4(A-Ce^{-x})+5(Ax+B+Ce^{-x})=10x+e^{-x}\\

Equating the coefficients

5Ax+2Ce^{-x}+4A+5B=10x+e^{-x}\\5A=10\\A=2\\4A+5B=0\\B=-\frac{4A}{5}B=-\frac{8}{5}2C=1\\C=\frac{1}{2}\\So,\\y_{p}=2x+\frac{1}{2}e^{-x}-\frac{8}{5}---(3)\\

The general solution is

y=y_{h}+y_{p}

from (2) ad (3)

y=e^{-2x}(c_{1}cos(x)+c_{2}sin(x))+2x+\frac{1}{2}e^{-x}-\frac{8}{5}

6 0
3 years ago
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