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1 year ago

6

At a restaurant, when a customer buys four pretzels the fifth pretzel is free. soft pretzels cost $3.90 each. you ordered 12 sof

t pretzels what is your mean(average) cost per pretzel

1 answer:

elixir [45]1 year ago

3 0

Answer:

($3.90)(12) - (2)($3.90) = 39.00

39.00/12 = 3.25 per pretzel

Step-by-step explanation:

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Biologists are analyzing soil to check for the number of worms and grubs in a wildlife preserve. Let the random variable W repre

Vladimir79 [104]

Answer:

the answer is μ=5.28 and σ=1.67

Step-by-step explanation:

$E(W) =0\times0.05+1\times0.06+2\times0.18+3\times0.35+4\times0.3+5\times0.05+6\times0.01\\=0.06+0.36+1.05+1.2+0.25+0.06\\=2.98$

$E(W^2)=10.34\\\rightarrow Var(W) = E(W^2)-E^2(W)\\=1.4596$

$E(G) = 2.3\\Var(G)=E(G^2)-E^2(G)\\=6.62-5.29\\=1.33$

So,

$\mu = E(W+G)=E(W)+E(G)=5.28$

$\sigma ^2 = Var (W+G) \\=Var(W)+Var(G)=2.79\\\sigma =1.67$

Therefore, the answer is μ=5.28 and σ=1.67

5 0

2 years ago

On a coordinate plane, triangle L M N is shown. Point L is at (2, 4), point M is at (negative 2, 1), and point N is at (negative

Crank

Perimeter is 8 + √(10) units.

<u>Step-by-step explanation:</u>

First we have to find the distance between the all the 3 points and then using the distances, we can find the sum of it, so that we will get the perimeter of the triangle.

The points are L(2,4) , M(-2,1) and N(-1,4)

Distance between the points can be found as,

√((x₂-x₁)² + (y₂-y₁)²)

Plugin the values in the given point, we will get LM as,

LM = √((-2-2)² + (1-4)²)

= √((-4)² + (-3)²)

= √(16 + 9) = √(25) = 5 units.

MN = √((-1-(-2))² + (4-1)²)

= √((-1+2))² + (4-1)²)

= √((1))² + (3)²)

= √(1 + 9) = √(10)

NL = √((2-(-1))² + (4-4)²)

= √((3)² + 0²) = √(9) = 3 units

Perimeter = LM + MN + NL = 5+ √(10) + 3 = 8 + √(10) units.

6 0

1 year ago

Read 2 more answers

Let C be the curve which is the union of two line segments, the first going from (0, 0) to (4, -3) and the second going from (4,

NISA [10]

∫4dy + 3dx

You can do this using Green's theorem. The line integral over the boundary of the closed triangle.
See the partial derivative of the partial derivative of 3 with respect to y = 0

You can see that now the line integral can be divided into three sides of triangle

hope this helps

6 0

2 years ago

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3

In-s [12.5K]

Answer:

a) There is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

c) There is a 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.3 minutes and a standard deviation of 3.3 minutes. This means that $\mu = 8.3, \sigma = 3.3$ .

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

We are working with a sample mean of 37 jets. So we have that:

$s = \frac{3.3}{\sqrt{37}} = 0.5425$

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

This probability is the pvalue of Z when $X = 8.65$ . So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{8.65 - 8.3}{0.5425}$

$Z = 0.65$

$Z = 0.65$ has a pvalue of 0.7422. This means that there is a 74.22% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is subtracted by the pvalue of Z when $X = 7.43$

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{7.43 - 8.3}{0.5425}$

$Z = -1.60$

$Z = -1.60$ has a pvalue of 0.0548.

There is a 1-0.0548 = 0.9452 = 94.52% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Total time of 320 minutes for 37 jets, so

$X = \frac{320}{37} = 8.65$

Total time of 275 minutes for 37 jets, so

$X = \frac{275}{37} = 7.43$

This probability is the pvalue of Z when $X = 8.65$ subtracted by the pvalue of Z when $X = 7.43$ .

So:

From a), we have that for $X = 8.65$ , we have $Z = 0.65$ , that has a pvalue of 0.7422.

From b), we have that for $X = 7.43$ , we have $Z = -1.60$ , that has a pvalue of 0.0548.

So there is a 0.7422 - 0.0548 = 0.6874 = 68.74% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes.

7 0

2 years ago

George took a nonstop flight from Dallas to

In-s [12.5K]

It is given in the question that,

George took a nonstop flight from Dallas to Los Angeles, a total flight distance of 1,233 miles. The plane flew at a speed of 460 miles per hour for the first 75 minutes of the flight and at a speed of 439 miles per hour for the remainder of the flight.

Let for x hours, the flight travelled with a speed of 439 miles per hour .

So we have,

$460* \frac{75}{60} + 439*x = 1233$

$575 + 439x =1233
\\
439x = 1233 - 575
\\
439x = 658
\\
x = \frac{658}{439} hours$

And to convert it in minutes, we have to multiply by 60. And on doing so, we will get

$x = \frac{658}{439}*60 =approx \ 90 \ minutes$

5 0

2 years ago