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4 years ago

11

A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks along the x-axis from the spotlight toward the bui

lding at a speed of 1.2 m/s, which is taken as the given dx/dt, how fast is the length of his shadow on the building decreasing when he is 4 m from the building?

1 answer:

tester [92]4 years ago

7 0

Answer:

(dy/dt) = - 0.45 m/s (the minus sign indicates that the height of the shadow decreases as the man moves towards the wall).

Step-by-step explanation:

An image of the problem is presented in the attached file to this solution. The image shows the exact moment when the 2m tall man is 4m away from the wall.

First of, when the man is 4m away from the wall, he is 8m away from the spotlight, that is, x = 8m

Using the concept of similar triangles, we can obtain the value of y at this point.

(x/2) = [(x+4)/y]

x = 8

(8/2) = (12/y)

y = 3m when x = 8m

From the image,

But generally, for x and y at any point,

(x/2) = (12/y)

xy = 24

Differentiating with respect to t

(d/dt)(xy) = (d/dt)12

y (dx/dt) + x (dy/dt) = 0

x (dy/dt) = - y (dx/dt)

At x = 8m, y = 3m, (dx/dt) = 1.2 m/s

8 (dy/dt) = - 3 (1.2)

(dy/dt) = - 3.6/8 = - 0.45 m/s

The minus sign indicates that the height of the shadow decreases as the man moves towards the wall.

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The mean amount purchased by a typical customer at Churchill's Grocery Store is $26.00 with a standard deviation of $6.00. Assum

Vadim26 [7]

Answer:

a) 0.0951

b) 0.8098

c) Between $24.75 and $27.25.

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 normally distributed samples are 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.

Central Limit Theorem

The Central Limit Theorem estabilishes 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.

In this problem, we have that:

$\mu = 26, \sigma = 6, n = 62, s = \frac{6}{\sqrt{62}} = 0.762$

(a)

What is the likelihood the sample mean is at least $27.00?

This is 1 subtracted by the pvalue of Z when X = 27. So

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

By the Central Limit Theorem

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

$Z = \frac{27 - 26}{0.762}$

$Z = 1.31$

$Z = 1.31$ has a pvalue of 0.9049

1 - 0.9049 = 0.0951

(b)

What is the likelihood the sample mean is greater than $25.00 but less than $27.00?

This is the pvalue of Z when X = 27 subtracted by the pvalue of Z when X = 25. So

X = 27

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

$Z = \frac{27 - 26}{0.762}$

$Z = 1.31$

$Z = 1.31$ has a pvalue of 0.9049

X = 25

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

$Z = \frac{25 - 26}{0.762}$

$Z = -1.31$

$Z = -1.31$ has a pvalue of 0.0951

0.9049 - 0.0951 = 0.8098

c)Within what limits will 90 percent of the sample means occur?

50 - 90/2 = 5

50 + 90/2 = 95

Between the 5th and the 95th percentile.

5th percentile

X when Z has a pvalue of 0.05. So X when Z = -1.645

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

$-1.645 = \frac{X - 26}{0.762}$

$X - 26 = -1.645*0.762$

$X = 24.75$

95th percentile

X when Z has a pvalue of 0.95. So X when Z = 1.645

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

$1.645 = \frac{X - 26}{0.762}$

$X - 26 = 1.645*0.762$

$X = 27.25$

Between $24.75 and $27.25.

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3 years ago

What is the slope of the line?

Vsevolod [243]

Answer:

2

Step-by-step explanation:

Slope (m) is equal to the change in the y-coordinate divided by the change in the x-coordinate.

The change in y= 6

The change in x= 3

6/3 =2

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3 years ago

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Before the pandemic cancelled sports, a baseball team played home games in a stadium that holds up to 50,000 spectators. When ti

spayn [35]

Answer:

(a) $D(x)=-2,500x+60,000$

(b) $R(x)=60,000x-2500x^2$

(c) x=12

(d)Optimal ticket price: $12

Maximum Revenue:$360,000

Step-by-step explanation:

The stadium holds up to 50,000 spectators.

When ticket prices were set at $12, the average attendance was 30,000.

When the ticket prices were on sale for $10, the average attendance was 35,000.

(a)The number of people that will buy tickets when they are priced at x dollars per ticket = D(x)

Since D(x) is a linear function of the form y=mx+b, we first find the slope using the points (12,30000) and (10,35000).

$\text{Slope, m}=\dfrac{30000-35000}{12-10}=-2500$

Therefore, we have:

$y=-2500x+b$

At point (12,30000)

$30000=-2500(12)+b\\b=30000+30000\\b=60000$

Therefore:

$D(x)=-2,500x+60,000$

(b)Revenue

$R(x)=x \cdot D(x) \implies R(x)=x(-2,500x+60,000)\\\\R(x)=60,000x-2500x^2$

(c)To find the critical values for R(x), we take the derivative and solve by setting it equal to zero.

$R(x)=60,000x-2500x^2\\R'(x)=60,000-5,000x\\60,000-5,000x=0\\60,000=5,000x\\x=12$

The critical value of R(x) is x=12.

(d)If the possible range of ticket prices (in dollars) is given by the interval [1,24]

Using the closed interval method, we evaluate R(x) at x=1, 12 and 24.

$R(x)=60,000x-2500x^2\\R(1)=60,000(1)-2500(1)^2=\$57,500\\R(12)=60,000(12)-2500(12)^2=\$360,000\\R(24)=60,000(24)-2500(24)^2=\$0$

Therefore:

Optimal ticket price:$12
Maximum Revenue:$360,000

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3 years ago

(2^8 ⋅ 3^−5 ⋅ 6^0)^−2 ⋅ 3 to the power of negative 2 over 2 to the power of 3, whole to the power of 4 ⋅ 2^28 (5 points)

Otrada [13]

Answer:

Step-by-step explanation:

(2^8 ⋅ 3^−5 ⋅ 6^0)^−2 ⋅ 3

(2^8 ⋅ 3^−5 ⋅ 6^0)^−6

(a^m)^n=a^m*n

So,

(2^8*-6) * (3^−5*-6) * (6^0*-6)

(2^-48) (3^30) (6^0)

a^0=1

So,

(2^-48) (3^30) (1)

<u>(2^-48) (3^30)</u>

I hope this helps you friend!

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3 years ago

Can someone help me asap thx

Rufina [12.5K]

Answer:

1st = 18

2nd = -29

3rd = -6

4th= 8

Step-by-step explanation:

hope it helps

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3 years ago

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