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2 years ago

7

Sadie runs a farm stand that sells blueberries and peaches. Each pound of blueberries sells for $3.25 and each pound of peaches

sells for $3. Sadie sold 3 times as many pounds of peaches as pounds of blueberries and she made $171.50 altogether. Determine the number of pounds of blueberries sold and the number of pounds of peaches sold.

1 answer:

fiasKO [112]2 years ago

6 0

Answer:

There were 14 pounds of blueberries sold and 42 pounds of peaches sold.

Step-by-step explanation:

Since Sadie sold 3 times as many pounds of peaches as pounds of blueberries, she sold more pounds of peaches, so if we multiply 3 by the number of pounds of blueberries sold, we will get the number of pounds of peaches sold, meaning pp equals 3b.

There were 14 pounds of blueberries sold and 42 pounds of peaches sold.

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Accrotime is a manufacturer of quartz crystal watches. Accrotime researchers have shown that the watches have an average life of

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Answer:

a

$P(X < 24 )= 21.186\%$

b

$x = 19.78 \ months$

Step-by-step explanation:

From the question we are told that

The mean is $\mu = 26 \ months$

The standard deviation is $\sigma = 4 \ months$

Generally 2 year is equal to 24 months

Generally the percentage of total production will the company expect to replace is mathematically represented as

$P(X < 24 )= P(\frac{X - \mu }{ \sigma} < \frac{24 - 26}{4} )$

Generally $\frac{X - \mu}{\sigma } =Z (The \ standardized \ value \ of \ X )$

$P(X < 24 )= P(Z < -0.8 )$

Generally from the z-table

$P(Z < -0.8) = 0.21186$

So

$P(X < 24 )= 0.21186$

Converting to percentage

$P(X < 24 )= 0.21186 * 100$

=> $P(X < 24 )= 21.186\%$

Generally the duration that should be the guarantee period if Accrotime does not want to make refunds on more than 6% is mathematically evaluated as

$P(X < x) = P(\frac{X - \mu }{\sigma} < \frac{x - 26}{4} )= 0.06$

=> $P(X < x) = P(Z < \frac{x - 26}{4} )= 0.06$

From the normal distribution table the z-score for 0.06 at the lower tail is

$z = -1.555$

So

$\frac{x - 26}{4} = -1.555$

=> $x = 19.78 \ months$

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2 years ago

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Step-by-step explanation:

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The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat

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Answer:

1) $L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

2) $T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

3) $T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

8 0

2 years ago