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4 years ago

13

Jenny ball bounced 8 times farther than clarks ball. The two balls bounced a total distance of 36 inches. How far did Jenny's ba

ll bounce?

1 answer:

Galina-37 [17]4 years ago

4 0

Jenny = 8x
Clark's = x

8x + x = 36
9x = 36
9x/9 = 36/9
x = 4

Jenny = 8 x 4 = 32

So Clark's ball went the distant of 4 and Jenny's ball went 32

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This is my hw and it is due at 7!

AlladinOne [14]

Answer:

1. 60% (decrease by 40%)

3. 30% (decrease by 70%)

5. 250% (Increase by 150%)

7. 40% (Decrease by 60%)

9. 171.4% (increase 71.4%)

11. 152.6% (Increase by 52.6%)

Step-by-step explanation:

1. 20x = 11 = 0.55 rounds to 0.6 => 60%

3. 56 x = 14 = 0.25 rounds to 0.3 => 30%

5. 18x = 45 = 2.5 => 250%

7. 126x = 48 = 0.38 rounds to 0.4 => 40%

9. 42x = 72 = 1.714 => 171.4%

11. 95x = 145 = 1.526 => 152.6%

8 0

3 years ago

The lengths of two sides of a triangle are given. describe the lengths possible for the third side.

vodomira [7]

Answer:

7m and 17m

Step-by-step explanation:

The length would be between 7m and 17m. This is because since the sum of two sides of a triangle must be greater than the third side. The sum of the third side and 6 must give us something greater than 12. This means the least the length can be is 7m. Since 6+12=18, the third side also has to be less than 18. This means the greatest the length can be is 17m.

3 0

2 years ago

The number of defective circuit boards coming off a soldering machine follows a Poisson distribution. During a specific ten-hour

Alexus [3.1K]

Answer:

a) the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

c) the required probability is 0.2000

Step-by-step explanation:

Given the data in the question;

During a specific ten-hour period, one defective circuit board was found.

Lets X represent the number of defective circuit boards coming out of the machine , following Poisson distribution on a particular 10-hours workday which one defective board was found.

Also let Y represent the event of producing one defective circuit board, Y is uniformly distributed over ( 0, 10 ) intervals.

f(y) = $\left \{ {{\frac{1}{b-a} }\\\ }} \right _0$ ; ( a ≤ y ≤ b ) $_{elsewhere$

= $\left \{ {{\frac{1}{10-0} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

f(y) = $\left \{ {{\frac{1}{10} }\\\ }} \right _0$ ; ( 0 ≤ y ≤ 10 ) $_{elsewhere$

Now,

a) the probability that it was produced during the first hour of operation during that period;

P( Y < 1 ) = $\int\limits^1_0 {f(y)} \, dy$

we substitute

= $\int\limits^1_0 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^1_0$

= $\frac{1}{10} [ 1 - 0 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the first hour of operation is $\frac{1}{10}$ or 0.1000

b) The probability that it was produced during the last hour of operation during that period.

P( Y > 9 ) = $\int\limits^{10}_9 {f(y)} \, dy$

we substitute

= $\int\limits^{10}_9 {\frac{1}{10} } \, dy$

= $\frac{1}{10} [y]^{10}_9$

= $\frac{1}{10} [ 10 - 9 ]$

= $\frac{1}{10}$ or 0.1000

Therefore, the probability that the defective board was produced during the last hour of operation is $\frac{1}{10}$ or 0.1000

c)

no defective circuit boards were produced during the first five hours of operation.

probability that the defective board was manufactured during the sixth hour will be;

P( 5 < Y < 6 | Y > 5 ) = P[ ( 5 < Y < 6 ) ∩ ( Y > 5 ) ] / P( Y > 5 )

= P( 5 < Y < 6 ) / P( Y > 5 )

we substitute

$= (\int\limits^{6}_5 {\frac{1}{10} } \, dy) / (\int\limits^{10}_5 {\frac{1}{10} } \, dy)$

$= (\frac{1}{10} [y]^{6}_5) / (\frac{1}{10} [y]^{10}_5)$

= ( 6-5 ) / ( 10 - 5 )

= 0.2000

Therefore, the required probability is 0.2000

4 0

3 years ago

John the car salesman makes a commission on selling cars, as a percentage of the price, which

Bess [88]

Answer:

<u>John's commission percentage on new cars is x and his commission percentage on used cars is 1.5x, therefore, John's commission percentage on new cars is 2% and his commission percentage on used cars is 3%.</u>

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

John's commission percentage on new cars = x

John's commission percentage on used cars = x + 50%x (His commission on used cars is 50% more than on new cars)

Sales on new cars = $80,000

Sales on used cars = $50,000

John's commission check = $3,100

2. If x represents the commission percentage on new cars, express the commission percentage on used cars in terms of x.

John's commission percentage on new cars = x

John's commission percentage on used cars = x + 50%x = 1.5x

Now we can solve for x, this way:

80,000x + 50,000 * 1.5x = 3,100

80,000x + 75,000x = 3,100

155,000x = 3,100

x = 3,100/155,000

x = 0.02 = 2% ⇒ 1.5x = 3%

<u>John's commission percentage on new cars is 2% and his commission percentage on used cars is 3%</u>

3 0

3 years ago

Tim will borrow $6200 at 12.5% APR. He will pay it back over 2 years. What will be his Monthly payment be?

AlekseyPX

<span>The correct answer is D) $293.32.

Explanation:
We use the formula P=A/D, where P is the payment amount, A is the amount borrowed, and D is the discount factor.

The discount factor is given by the formula D={[(1+r)^n]-1}/[i(1+i)^n], where i is the monthly interest rate as a decimal number and n is the number of months taken for repayment.

For this problem, we have 12.5%; 12.5%=12.5/100=0.125.
This makes the monthly interest rate, i, 0.125/12=0.01042.

The number of months for repayment, n, will be 2*12=24.

Using these we have D={[1+0.01042)^24]-1}/[0.01042(1+0.01042)^24], which gives us D=21.1375.

We plug this in for D in our payment formula.

Additionally, we know that A=6200, since that is what is borrowed: P=6200/21.1375=293.32.</span>

3 0

3 years ago

Read 2 more answers