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4 years ago

8

angel is picking mountain blueberries for a delicious pie. she can pick 1/6 cup of blueberries in 2 minutes. if she needs 2 1/2

cups of blueberries for the pie, how long will it take her to to pick the berries?

1 answer:

Vadim26 [7]4 years ago

3 0

I would take her 30 minutes

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Distributive property of 7(5j + 8 +3j)

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Answer:

35j+56+21j

Step-by-step explanation:

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If a student knows a car traveled 300 kilometers in three hours, what can the student find?

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Its speed 100 km per hour. 300/3

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Suppose you want to eat lunch at a popular restaurant. The restaurant does not take reservations, so there is usually a waiting

dimulka [17.4K]

Answer:

a) $P(X>20 |X>15) =\frac{P(X>20)}{P(X>15)} =\frac{0.305}{0.742}=0.411$

b) $P(X>25 |X>18) =\frac{P(X>25)}{P(X>18)} =\frac{0.048}{0.481}=0.0997$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent the interpupillary distance (the distance between the pupils of the left and right eyes) of a population, and for this case we know the distribution for X is given by:

$X \sim N(17.8,4.3)$

Where $\mu=17.8$ and $\sigma=4.3$

And let $\bar X$ represent the sample mean, the distribution for the sample mean is given by:

$\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})$

On this case $\bar X \sim N(72,\frac{6}{\sqrt{27}})$

Part a

(a) What is the probability that the waiting time will exceed 20 minutes, given that it has exceeded 15 minutes?

$P(X>20 |X>15) =\frac{P(X>20 and X>15)}{P(X>15)}=\frac{P(X>20)}{P(X>15)}=\frac{1-P(X$

We can find the inidivual probabilities like this:

$P(X>15)=1-P(X$

$P(X>20)=1-P(X$

And then we can replace

$P(X>20 |X>15) =\frac{P(X>20)}{P(X>15)} =\frac{0.305}{0.742}=0.411$

Part b

(b) What is the probability that the waiting time will exceed 25 minutes, given that it has exceeded 18 minutes?

$P(X>25 |X>18) =\frac{P(X>25 and X>18)}{P(X>18)}=\frac{P(X>25)}{P(X>18)}=\frac{1-P(X$

We can find the inidivual probabilities like this:

$P(X>18)=1-P(X$

$P(X>25)=1-P(X$

And then we can replace

$P(X>25 |X>18) =\frac{P(X>25)}{P(X>18)} =\frac{0.048}{0.481}=0.0997$

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3 years ago

1. how can we get equation B from A<br><br>A. 4x+2=6-x<br><br>B. 5x+2=6

Fynjy0 [20]

Answer:

since equation a has -x on one side just add x to both sides to get rid of the -x and since you do it to both sides you maintain balance

Step-by-step explanation:

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A homeowner plans to enclose a 200 square foot rectangular playground in his garden, with one side along the boundary of his pro

Anestetic [448]

Answer:

Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.

Step-by-step explanation:

Given that, a homeowner plans to enclose a 200 square foot rectangle playground.

Let the width of the playground be y and the length of the playground be x which is the side along the boundary.

The perimeter of the playground is = 2(length +width)

=2(x+y) foot

The material costs $1 per foot.

Therefore total cost to give boundary of the play ground

=$[ 2(x+y)×1]

=$[2(x+y)]

But the neighbor will play one third of the side x foot.

So the neighbor will play $=\$(\frac13x)$

Now homeowner's total cost for the material is

$=\$[ 2(x+y)-\frac13x]$

$=\$[2x+2y-\frac13x]$

$=\$[2y+x+x-\frac13x]$

$=\$[2y+x+\frac{3x-x}{3}]$

$=\$[2y+x+\frac{2}{3}x]$

$=\$[2y+\frac53x]$

$\therefore C(x)=2y+\frac53x$

where C(x) is total cost of material in $.

Given that the area of the playground is 200.

We know that,

The area of a rectangle is =length×width

=xy square foot

∴xy=200

$\Rightarrow y=\frac{200}{x}$

Putting the value of y in C(x)

$\therefore C(x)=2(\frac{200}{x})+\frac53x$

The domain of C is $(0,\infty )$ .

$\therefore C(x)=2(\frac{200}{x})+\frac53x$

Differentiating with respect to x

$C'(x)= - \frac{400}{x^2}+\frac53$

Again differentiating with respect to x

$C''(x) = \frac{800}{x^3}$

To find the critical point set C'(x)=0

$\therefore 0= - \frac{400}{x^2}+\frac53$

$\Rightarrow \frac{400}{x^2}=\frac{5}{3}$

$\Rightarrow x^2 =\frac{400\times 3}{5}$

$\Rightarrow x=\sqrt{240}$

$\Rightarrow x=15.49 \approx15.5$

Therefore

$\left C''(x) \right|_{x=15.5}=\frac{800}{15.5^3}>0$

Therefore at x= 15.5 , C(x) is minimum.

Putting the value of x in $y=\frac{200}{x}$ we get

$\therefore y=\frac{200}{15.5}$

=12.9

Therefore the length and width of the playground are 15.5 feet and 12.9 feet respectively.

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4 years ago