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3 years ago

10

A kite flying in the air has a 12 line attached to it. Its line is pulled taut and casts a 9 shadow. Find the height of the kite

. If necessary, round your answer to the nearest tenth.

1 answer:

alexira [117]3 years ago

4 0

Answer:

8 (7.94)

Step-by-step explanation:

You can think of it as a geometry problem.

What is formed here is a triangle, which sides ate: the line, the line's shadow, and the height from the ground to the kite (here I attach a drawing).

What you need to find is the height. We will call it H.

As the triangle formed is a right one, we can use Pitágoras' theorem. The height H squared plus the squared of the shadow is equal to the squared of the line (the hypotenuse). This is:

H^2 + 9^2 = 12^2

H^2 + 81= 144

H^2 = 63

Applying squared root in both sides

H = √63

H = 7,94

So, the height is approximately 8.

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Donna sent her grandson a birthday present. The box weighed 0.875 kilograms and the present itself weighed 6,800 grams. a. What

Nana76 [90]

A) We know that the box weighed 0.875 kg, and the present weighed 6,800 g. However we need the gross weight of the package in kg, so we have to convert the present's weight to kg. There are 1000 grams in 1 kg, so we can divide the present's weight by 100 to find the weight in kg:

$\frac{6,800}{1000} =6.8 kg$

Now that we have converted to grams, let's add the weight to get the total gross weight:

$0.875+6.8=7.675 kg$

So the gross weight of the package is 7.675 kg.

b) Shipping costs 74 cents per kg, so in order to find the total cost of shipping for the package, let's multiply the weight of the package by the cost of shipping:

$\frac{7.675kg}{1}* \frac{74 cents}{kg}=567.95 cents$

However the cost of shipping has to be in dollars. There are 100 cents in one dollar, so we need to divide the amount of cents that shipping costs by 100 to get the total in dollars:

$\frac{567.95cents}{100} =5.68dollars$

Now we know that the total cost of shipping, in dollars, is equal to $5.68.

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3 years ago

Amy and Mary are playing a board game that uses game money if $5 and $10. At the end of the game, Amy has n bills worth $5 each

loris [4]

Answer:

5n + 10d represented the money spend by the Amy in playing board games of cost $5 and $10 .

Step-by-step explanation:

As given

Amy and Mary are playing a board game that uses game money if $5 and $10.

At the end of the game, Amy has n bills worth $5 each and d bills worth $10 each .

Total amount of n bills = 5n

Total amount of d bills = 10d

Thus

Total money spend by the Amy in playing board games = Total amount of n bills + Total amount of d bills

= 5n + 10d

Therefore 5n + 10d represented the money spend by the Amy in playing board games of cost $5 and $10 .

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3 years ago

The height, h, in feet of a rock above the ground is given by the equation h = -16t^2 + 24t + 16, where t is the time in seconds

boyakko [2]

The rock will hit the ground at 2 seconds.

Explanation:

Given that the height, h, in feet of a rock above the ground is given by the equation $h=-16t^2+24t+16$ , where t is the time in seconds after the rock is thrown.

We need to determine the time at which the rock will hit the ground.

To determine the time, let us equate the height h = 0 in the equation $h=-16t^2+24t+16$ , we get,

$0=-16t^2+24t+16$

Switch sides, we have,

$-16 t^{2}+24 t+16=0$

Let us solve the equation using the quadratic formula.

Thus, we have,

$t=\frac{-24 \pm \sqrt{24^{2}-4(-16) 16}}{2(-16)}$

Simplifying, we get,

$t=\frac{-24 \pm \sqrt{576+1024}}{-32}$

$t=\frac{-24 \pm \sqrt{1600}}{-32}$

$t=\frac{-24 \pm 40}{-32}$

Hence, the two values of t are

$t=\frac{-24 + 40}{-32}$ and $t=\frac{-24 -40}{-32}$

Simplifying the values, we get,

$t=\frac{16}{-32}$ and $t=\frac{-64}{-32}$

Dividing, we get,

$t=-\frac{1}{2}$ and $t=2$

Since, t cannot negative values, we have,

$t=2$

Thus, the rock will hit the ground at 2 seconds.

8 0

3 years ago

A salesperson contacts eight potential customers per day. From past experience, we know that the probability of a potential cust

AlexFokin [52]

Answer:

(a) The probability the salesperson will make exactly two sales in a day is 0.1488.

(b) The probability the salesperson will make at least two sales in a day is 0.1869.

(c) The percentage of days the salesperson does not makes a sale is 43.05%.

(d) The expected number of sales per day is 0.80.

Step-by-step explanation:

Let <em>X</em> = number of sales made by the salesperson.

The probability that a potential customer makes a purchase is 0.10.

The salesperson contacts <em>n</em> = 8 potential customers per day.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.

The probability mass function of <em>X</em> is:

$P(X=x)={8\choose x}0.10^{x}(1-0.10)^{8-x};\ x=0,1,2,3...$

(a)

Compute the probability the salesperson will make exactly two sales in a day as follows:

$P(X=2)={8\choose 2}0.10^{2}(1-0.10)^{8-2}\\=28\times 0.01\times 0.5314\\=0.1488$

Thus, the probability the salesperson will make exactly two sales in a day is 0.1488.

(b)

Compute the probability the salesperson will make at least two sales in a day as follows:

P (X ≥ 2) = 1 - P (X < 2)

= 1 - P (X = 0) - P (X = 1)

$=1-{8\choose 0}0.10^{0}(1-0.10)^{8-0}-{8\choose 1}0.10^{1}(1-0.10)^{8-1}\\=1-0.4305-0.3826\\=0.1869$

Thus, the probability the salesperson will make at least two sales in a day is 0.1869.

(c)

Compute the probability that a salesperson does not makes a sale is:

$P(X=0)={8\choose 0}0.10^{0}(1-0.10)^{8-0}\\=8\times 1\times 0.4305\\=0.4305$

The percentage of days the salesperson does not makes a sale is,

0.4305 × 100 = 43.05%

Thus, the percentage of days the salesperson does not makes a sale is 43.05%.

(d)

Compute the expected number of sales per day as follows:

$E(X)=np=8\times 0.10=0.80$

Thus, the expected number of sales per day is 0.80.

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3 years ago

If a semi truck's trailer is 40 feet long, 10 feet wide, and has a height of 8 feet, what is its volume?

Temka [501]

First multiply 40 and 10 to get 400. Then multiply 400 by 8 to get 3,200.

Your answer would be: 3,200

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3 years ago