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Blababa [14]
2 years ago
6

Sue had a circular picture that she wanted to frame. If the picture's frame diameter was

Mathematics
1 answer:
sesenic [268]2 years ago
7 0
Sue had 7.5 inches framing
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A ratio of dogs to cats a pet store is 1:3. There are 15 cats. how many dogs are at the pet store
DIA [1.3K]

From the ratio given we know that there is 3 times more cats than dogs.

If there are 15 cats, then there are 15/3 dogs or 5 dogs.

answer: 5 dogs

8 0
3 years ago
there are 24 apple trees in my orchard. if i can harvest the fruit from 2 and 2 thirds trees each day, how many days will it tak
valentina_108 [34]

Answer: 9 days.

Step-by-step explanation:

2 2/3 is the amount able to be harvested, you want to find days. So, 2 2/3(d)=9. You need to solve for days (d) which equals 9.

8 0
3 years ago
How do you solve 4(8x+5)
alexandr1967 [171]

Answer:

32x+20

Step-by-step explanation:

4*8x+4*5=

32x+20

3 0
3 years ago
Read 2 more answers
Write the equation of the line that passes through (−3,1) and (2,−1) in slope-intercept form
Alex787 [66]

Answer:

y=-\frac{2}{5}x-\frac{1}{5}

Step-by-step explanation:

The equation of a line is y = mx + b

Where:

  • m is the slope
  • b is the y-intercept

First, let's find what m is, the slope of the line.

Let's call the first point you gave, (-3,1), point #1, so the x and y numbers given will be called x1 and y1.

Also, let's call the second point you gave, (2,-1), point #2, so the x and y numbers here will be called x2 and y2.

Now, just plug the numbers into the formula for m above, like this:

m = -\frac{2}{5}

So, we have the first piece to finding the equation of this line, and we can fill it into y=mx+b like this:

y=-\frac{2}{5}x + b

Now, what about b, the y-intercept?

To find b, think about what your (x,y) points mean:

  • (-3,1). When x of the line is -3, y of the line must be 1.
  • (2,-1). When x of the line is 2, y of the line must be -1.

Now, look at our line's equation so far: y=-\frac{2}{5}x + b. b is what we want, the --\frac{2}{5} is already set and x and y are just two 'free variables' sitting there. We can plug anything we want in for x and y here, but we want the equation for the line that specfically passes through the two points (-3,1) and (2,-1).

So, why not plug in for x and y from one of our (x,y) points that we know the line passes through? This will allow us to solve for b for the particular line that passes through the two points you gave!

You can use either (x,y) point you want. The answer will be the same:

  • (-3,1). y = mx + b or 1=-\frac{2}{5} * -3 + b, or solving for b: b = 1-(-\frac{2}{5})(-3).b = -\frac{1}{5}.
  • (2,-1). y = mx + b or -1=-\frac{2}{5} * 2 + b, or solving for b: b = 1-(-\frac{2}{5})(2). b = -\frac{1}{5}.

See! In both cases, we got the same value for b. And this completes our problem.

The equation of the line that passes through the points  (-3,1) and (2,-1) is y=-\frac{2}{5}x-\frac{1}{5}

8 0
3 years ago
The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9
Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% 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:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 8.9 minutes and a standard deviation of 2.9 minutes.

This means that \mu = 8.9, \sigma = 2.9

Sample of 37:

This means that n = 37, s = \frac{2.9}{\sqrt{37}}

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

320/37 = 8.64865

Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{8.64865 - 8.9}{\frac{2.9}{\sqrt{37}}}

Z = -0.53

Z = -0.53 has a p-value of 0.2981

0.2981 = 29.81% 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?

275/37 = 7.4324

Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So

Z = \frac{X - \mu}{s}

Z = \frac{7.4324 - 8.9}{\frac{2.9}{\sqrt{37}}}

Z = -3.08

Z = -3.08 has a p-value of 0.001

1 - 0.001 = 0.999

0.999 = 99.9% 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?

Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So

0.2981 - 0.0010 = 0.2971

0.2971 = 29.71% 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
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