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3 years ago

Using the following table. What is the probability of selecting a sandwich and chip combination that has BBQ chips?

Mathematics

1 answer:

snow_lady [41]3 years ago

6 0

Answer:

1/7

Step-by-step explanation:

1/7×1/7×1/7×1/7=

1/7

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Find the amplitude, period, and phase shift of the function.

Softa [21]

Cosine is transformed in the form:

$f(x)=a\cos(b(x-c))+d$

The amplitude is a

The period of cosine is:

$\frac{2\pi}{b}$

And the phase shift is always by c, so that means our transformed functon in the form mentioned above looks like:

$f(x)=3\cos{(3(x-\frac{\pi}{6}}))+1$

So our period is $\frac{2\pi}{3}$

Our phase shift is by $\frac{\pi}{6}$ units to the right

and our amplitude is 3.

4 0

2 years ago

Parker drew a triangle with no cogruent sides and a 95° angle. Classify the triangle by the lengths of its sides and measures of

ss7ja [257]

Obtuse scalene, no equal sides, so no equal angles, and the one angle is over 90°

6 0

2 years ago

PLEASE HELP IM TIMED

rewona [7]

Answer:

Graph D.

Step-by-step explanation:

The line goes through the origin, and therefore shows direct variation.

6 0

3 years ago

Read 2 more answers

A healthcare provider monitors the number of CAT scans performed each month in each of its clinics. The most recent year of data

Rasek [7]

Answer: a. 1.981 < μ < 2.18

b. Yes.

Step-by-step explanation:

A. For this sample, we will use t-distribution because we're estimating the standard deviation, i.e., we are calculating the standard deviation, and the sample is small, n = 12.

First, we calculate mean of the sample:

$\overline{x}=\frac{\Sigma x}{n}$

$\overline{x}=\frac{2.31=2.09+...+1.97+2.02}{12}$

$\overline{x}=$ 2.08

Now, we estimate standard deviation:

$s=\sqrt{\frac{\Sigma (x-\overline{x})^{2}}{n-1} }$

$s=\sqrt{\frac{(2.31-2.08)^{2}+...+(2.02-2.08)^{2}}{11} }$

s = 0.1564

For t-score, we need to determine degree of freedom and $\frac{\alpha}{2}$ :

df = 12 - 1

df = 11

$\alpha$ = 1 - 0.95

α = 0.05

$\frac{\alpha}{2}=$ 0.025

Then, t-score is

$t_{11,0.025}$ = 2.201

The interval will be

$\overline{x}$ ± $t.\frac{s}{\sqrt{n} }$

2.08 ± $2.201\frac{0.1564}{\sqrt{12} }$

2.08 ± 0.099

The 95% two-sided CI on the mean is 1.981 < μ < 2.18.

B. We are 95% confident that the true population mean for this clinic is between 1.981 and 2.18. Since the mean number performed by all clinics has been 1.95, and this mean is less than the interval, there is evidence that this particular clinic performs more scans than the overall system average.

8 0

2 years ago

A golfball is hit from the ground with an initial velocity of 200 ft/sec. The horizontal distance that the golfball will travel,

algol13

Answer:

The golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

Step-by-step explanation:

The formula from the maximum distance of a projectile with initial height h=0, is:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}$

Where $v_i$ is the initial velocity.

In the closed interval method, the first step is to find the values of the function in the critical points in the interval which is $[0, \pi/2]$ . The critical points of the function are those who make $d'(\theta)=0$ :

$d(\theta)=\frac{v_i^2\sin(2\theta)}{g}\\d'(\theta)=\frac{v_i^2\cos(2\theta)}{g}*(2)\\d'(\theta)=\frac{2v_i^2\cos(2\theta)}{g}$

$d'(\theta)=0\\\frac{2v_i^2\cos(2\theta)}{g}=0\\\cos(2\theta)=0\\2\theta=\pi/2,3\pi/2,5\pi/2,...\\\theta=\pi/4,3\pi/4,5\pi/4,...$

The critical value inside the interval is $\pi/4$ .

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\d(\pi/4)=\frac{v_i^2sin(2(\pi/4))}{g}\\d(\pi/4)=\frac{v_i^2sin(\pi/2)}{g}\\d(\pi/4)=\frac{v_i^2(1)}{g}\\d(\pi/4)=\frac{(200)^2}{32}\\d(\pi/4)=\frac{40000}{32}\\d(\pi/4)=1250ft$

The second step is to find the values of the function at the endpoints of the interval:

$d(\theta)=\frac{v_i^2sin(2\theta)}{g}\\\theta=0\\d(0)=\frac{v_i^2sin(2(0))}{g}\\d(0)=\frac{v_i^2(0)}{g}=0ft\\\theta=\pi/2\\d(\pi/2)=\frac{v_i^2sin(2(\pi/2))}{g}\\d(\pi/2)=\frac{v_i^2sin(\pi)}{g}\\d(\pi/2)=\frac{v_i^2(0)}{g}=0ft$

The biggest value of f is gived by $\pi/4$ , therefore $\pi/4$ is the absolute maximum.

In the context of the problem, the golfball launched with an initial velocity of 200ft/s will travel the maximum possible distance which is 1250 ft when it is hit at an angle of $\pi/4$ .

4 0

2 years ago