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MariettaO [177]
3 years ago
13

You throw a ball at a height of 5 feet above the ground. The height h (in feet) of the ball after t seconds can be modeled by th

e equation, h=(-16t)^2 + 44 + 5.
The question is:
After how many seconds does the ball reach a height of 15 feet?
WILL GIVE BRAINLIEST IF ANSWERED CORRECTLY
Mathematics
1 answer:
marusya05 [52]3 years ago
3 0

Answer:?

Step-by-step explanation:

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Suppose PR = 54, solve for QR<br> 4x-1<br> 3x-1<br> P<br> R
Svet_ta [14]

Answer:

QR = 23

Step-by-step explanation:

P, A, and R are collinear.

PR = 54

PQ = 4x - 1

QR = 3x - 1

To solve for the numerical length of PR, let's generate an equation to find the value of x.

According to the segment addition postulate:

PQ + QR = PR

(4x - 1) + (3x - 1) = 54 (substitution)

Solve for x

4x - 1 + 3x - 1 = 54

Combine like terms

4x + 3x - 1 - 1 = 54

7x - 2 = 54

Add 2 to both sides

7x - 2 + 2 = 54 + 2

7x = 56

Divide both sides by 7

\frac{7x}{7} = \frac{56}{7}

x = 8

QR = 3x - 1

Plug in the value of x into the equation

QR = 3(8) - 1 = 24 - 1

QR = 23

5 0
3 years ago
A distance AB is observed repeatedly using the same equipment and procedures, and the results, in meters, are listed below: 67.4
skad [1K]

Answer:

a) \bar X =\frac{67.401+67.400+67.402+67.396+67.406+67.401+67.396+67.401+67.405+67.404}{10}=67.401

b) The sample deviation is calculated from the following formula:

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And for this case after replace the values and with the sample mean already calculated we got:

s=0.0036

If we assume that the data represent a population then the standard deviation would be given by:

\sigma=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n}}

And then the deviation would be:

\sigma =0.00319

Step-by-step explanation:

For this case we have the following dataset:

67.401, 67.400, 67.402, 67.396, 67.406, 67.401, 67.396, 67.401, 67.405, and 67.404

Part a: Determine the most probable value.

For this case the most probably value would be the sample mean given by this formula:

\bar X =\frac{\sum_{i=1}^n X_i}{n}

And if we replace we got:

\bar X =\frac{67.401+67.400+67.402+67.396+67.406+67.401+67.396+67.401+67.405+67.404}{10}=67.401

Part b: Determine the standard deviation

The sample deviation is calculated from the following formula:

s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}

And for this case after replace the values and with the sample mean already calculated we got:

s=0.0036

If we assume that the data represent a population then the standard deviation would be given by:

\sigma=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n}}

And then the deviation would be:

\sigma =0.00319

3 0
4 years ago
Jenny and her children went to the SeaWorld.She purchased adult ticket for herself and children tickets for her 2 boys. Tickets
dangina [55]

Given:

She purchased adult ticket for herself and children tickets for her 2 boys.

Cost of one adult ticket = $39.99

Cost of one child ticket = $19.99

To find:

The expression to represent the total cost of the tickets Jenny purchased.

Solution:

Let x represent the number of adult ticket purchase

and y represent the number of child tickets purchase.

Cost of one adult ticket = $39.99

Cost of one child ticket = $19.99

Total cost = 39.99x + 19.99y

Therefore, the expression for total cost is 39.99x + 19.99y.

She purchased one adult ticket and 2 child tickets.

Substitute x=1 and y=2 in the above expression.

Total cost = 39.99(1) + 19.99(2)

               = 39.99 + 39.98

               = 79.97

Therefore, the required expression to represent the total cost of the tickets Jenny purchased is 39.99(1) + 19.99(2) and total cost is $79.97.

3 0
3 years ago
Movie Ticket Sales : 6x + 9y = 1,500 Suppose on this day all tickets were on sale for the child price of $6. How many tickets we
dedylja [7]
The answer to the question

6 0
3 years ago
Read 2 more answers
Al bob and carl can life 240 pounds.
padilas [110]

Answer:

1020 POUNDS.

Step-by-step explanation:

i think im not sure if i am wrong tell me and i wiil redo! :)

5 0
3 years ago
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