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3 years ago

When a 6-ft student casts a 17-ft shadow, a tree casts a shadow that is 102 ft long. Find the height of the tree.

Mathematics

1 answer:

lukranit [14]3 years ago

8 0

This is what i got :) hope it helps

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A system of equations consists of two lines. One line passes through (-1, 3) and (0, 1). The other line passes through (1, 4) an

Setler79 [48]

The equation of the line in its generic form is:
y = mx + b
Where,
m = (y2-y1) / (x2-x1)

For (-1, 3) and (0, 1):
We look for the value of m:
m = (1-3) / (0 - (- 1))
m = (- 2) / (0 + 1)
m = -2
We look for the value of b:
1 = m (0) + b
b = 1
The line is:
y = -2x + 1

For (1, 4) and (0, 2):
We look for the value of m:
m = (2-4) / (0-1)
m = (- 2) / (- 1)
m = 2
We look for the value of b:
2 = m (0) + b
b = 2
The line is:
y = 2x + 2

The system of equations is:
y = -2x + 1
y = 2x + 2

Answer:
the system has one solution

4 0

3 years ago

The Acme Company manufactures widgets. The distribution of widget weights is bell-shaped. The widget weights have a mean of 55 o

Usimov [2.4K]

Answer:

a) For this case and using the empirical rule we can find the limits in order to have 9% of the values:

$\mu -2\sigma = 55 -2*6 =43$

$\mu +2\sigma = 55 +2*6 =67$

95% of the widget weights lie between 43 and 67

b) For this case we know that 37 is 3 deviations above the mean and 67 2 deviations above the mean since within 3 deviation we have 99.7% of the data then the % below 37 would be (100-99.7)/2 = 0.15% and the percentage above 67 two deviations above the mean would be (100-95)/2 =2.5% and then we can find the percentage between 37 and 67 like this:

$100 -0.15-2.5 = 97.85$

c) We want to find the percentage above 49 and this value is 1 deviation below the mean so then this percentage would be (100-68)/2 = 16%

Step-by-step explanation:

For this case our random variable of interest for the weights is bell shaped and we know the following parameters.

$\mu = 55, \sigma =6$

We can see the illustration of the curve in the figure attached. We need to remember that from the empirical rule we have 68% of the values within one deviation from the mean, 95% of the data within 2 deviations and 99.7% of the values within 3 deviations from the mean.

Part a

For this case and using the empirical rule we can find the limits in order to have 9% of the values:

$\mu -2\sigma = 55 -2*6 =43$

$\mu +2\sigma = 55 +2*6 =67$

95% of the widget weights lie between 43 and 67

Part b

For this case we know that 37 is 3 deviations above the mean and 67 2 deviations above the mean since within 3 deviation we have 99.7% of the data then the % below 37 would be (100-99.7)/2 = 0.15% and the percentage above 67 two deviations above the mean would be (100-95)/2 =2.5% and then we can find the percentage between 37 and 67 like this:

$100 -0.15-2.5 = 97.85$

Part c

We want to find the percentage above 49 and this value is 1 deviation below the mean so then this percentage would be (100-68)/2 = 16%

4 0

4 years ago

For each week of work, Aaron gets paid a base amount of $500 plus 25% of his total sales. If y represents his total pay each wee

Allisa [31]

Answer: y = $500 + (25%)x

Y = $500 plus 25% times x

Step-by-step explanation:

4 0

3 years ago

Solve the above que no. 55

aleksandr82 [10.1K]

Answer:

Let $\left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right)$ , we proceed to prove the trigonometric expression by trigonometric identity:

1) $\left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right)$ Given

2) $\left(1+\frac{\cos^{2}A}{\sin^{2}A} \right)\cdot \left(1+\frac{\sin^{2}A}{\cos^{2}A} \right)$ $\tan A = \frac{1}{\cot A} = \frac{\sin A}{\cos A}$

3) $\left(\frac{\sin^{2}A+\cos^{2}A}{\sin^{2}A} \right)\cdot \left(\frac{\cos^{2}A+\sin^{2}A}{\cos^{2}A} \right)$

4) $\left(\frac{1}{\sin^{2}A} \right)\cdot \left(\frac{1}{\cos^{2}A} \right)$ $\sin^{2}A+\cos^{2}A = 1$

5) $\frac{1}{\sin^{2}A\cdot \cos^{2}A}$

6) $\frac{1}{\sin^{2}A\cdot (1-\sin^{2}A)}$ $\sin^{2}A+\cos^{2}A = 1$

7) $\frac{1}{\sin^{2}A-\sin^{4}A}$ Result

Step-by-step explanation:

Let $\left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right)$ , we proceed to prove the trigonometric expression by trigonometric identity:

1) $\left(1+\frac{1}{\tan^{2}A} \right)\cdot \left(1+\frac{1}{\cot^{2}A} \right)$ Given

2) $\left(1+\frac{\cos^{2}A}{\sin^{2}A} \right)\cdot \left(1+\frac{\sin^{2}A}{\cos^{2}A} \right)$ $\tan A = \frac{1}{\cot A} = \frac{\sin A}{\cos A}$

3) $\left(\frac{\sin^{2}A+\cos^{2}A}{\sin^{2}A} \right)\cdot \left(\frac{\cos^{2}A+\sin^{2}A}{\cos^{2}A} \right)$

4) $\left(\frac{1}{\sin^{2}A} \right)\cdot \left(\frac{1}{\cos^{2}A} \right)$ $\sin^{2}A+\cos^{2}A = 1$

5) $\frac{1}{\sin^{2}A\cdot \cos^{2}A}$

6) $\frac{1}{\sin^{2}A\cdot (1-\sin^{2}A)}$ $\sin^{2}A+\cos^{2}A = 1$

7) $\frac{1}{\sin^{2}A-\sin^{4}A}$ Result

4 0

3 years ago

What is the probability of rolling a number less than five on a standard number cube (6 sides)

fgiga [73]

<h3>Answer: 2/3</h3>

Here are the outcomes that are less than five = {1, 2, 3, 4}. There are A = 4 values here.

Here are all the outcomes possible: {1, 2, 3, 4, 5, 6}. There are B = 6 ways to roll a single number cube (aka dice).

Divide the two values: A/B = 4/6 = (2*2)/(2*3) = 2/3

The fraction 2/3 is approximately equal to 0.667 which converts to 66.7%

4 0

3 years ago