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Aloiza [94]
3 years ago
8

Help me Please Please Please Please Pleaseeeeeeee

Mathematics
1 answer:
Thepotemich [5.8K]3 years ago
6 0
H = 151t - 16t²

The height of the ball when it return to the ground will be 0
0 = 151t - 16t²

The zero product property is that when two numbers are being multiplied and the product is 0, one of them must be equal to 0. Therefore, we can factorize this equation:
16t² - 151t = 0
t(16t - 151t) = 0
By the zero product property:
t = 0   or  16t - 151 = 0

So t = 0 or t = 9.44 seconds

The first solution is before he releases the ball and the second is when the ball comes back to the ground. Thus, the ball's air time is 9.44 seconds.
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The automaker of your new car recommends an oil change following the first 5,000 miles and then every 3,000 miles thereafter. If
Jlenok [28]

To determine how many oil changes you will need the first year, you can calculate the number of miles driven the first year and then use that to determine how many groups of 3000 miles you will drive after the first 5000 miles.


1. 52 x 250 = 13000 miles

2. 13000 miles - 5000 miles (1 oil change) = 8000 miles

3. 8000 miles - 3000 miles (1 oil change) - 3000 miles (1 oil change) = 2000 miles


The first year, you will need 3 oil changes.

7 0
3 years ago
7x+39>=53. And. 16x+ 15>31 X>=2 x>1. X<=2. No solution
bazaltina [42]

Answer:

7x+39>=53 and 16x+ 15>31 x>=2 (no solutions exist)

Step-by-step explanation:

4 0
3 years ago
The wind blew away 33 of your muffins. That was 3/4 of all of them! With how many did you start
Vedmedyk [2.9K]

Answer:

44

because each that's ¼ has 11 so 11x4 =44

6 0
3 years ago
Read 2 more answers
Let represent the number of tires with low air pressure on a randomly chosen car. The probability distribution of is as follows.
Sindrei [870]

Answer:

a) P(X=3) = 0.1

b) P(X\geq 3) =1-P(X

And replacing we got:

P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4

c) P(X=4) = 0.3

d) P(X=0) = 0.2

e) E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2

f) E(X^2)= \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4

And the variance would be:

Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4

And the deviation:

\sigma =\sqrt{2.4} = 1.549

Step-by-step explanation:

We have the following distribution

x      0     1     2   3   4

P(x) 0.2 0.3 0.1 0.1 0.3

Part a

For this case:

P(X=3) = 0.1

Part b

We want this probability:

P(X\geq 3) =1-P(X

And replacing we got:

P(X \geq 3) = 1- [0.2+0.3+0.1]= 0.4

Part c

For this case we want this probability:

P(X=4) = 0.3

Part d

P(X=0) = 0.2

Part e

We can find the mean with this formula:

E(X)= \sum_{i=1}^n X_i P(X_i)

And replacing we got:

E(X) =0*0.2 +1*0.3+2*0.1 +3*0.1 +4*0.3= 2

Part f

We can find the second moment with this formula

E(X^2)= \sum_{i=1}^n X^2_i P(X_i)

And replacing we got:

E(X^2) =0^2*0.2 +1^2*0.3+2^2*0.1 +3^2*0.1 +4^2*0.3= 6.4

And the variance would be:

Var(X0 =E(X^2)- [E(X)]^2 = 6.4 -(2^2)= 2.4

And the deviation:

\sigma =\sqrt{2.4} = 1.549

4 0
4 years ago
What's is 2/5 of 20<br> A. 4<br> B. 8<br> C. 20 2/5<br> D. 40
steposvetlana [31]
2/5 x 20 = 40/5 = B 8
5 0
3 years ago
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