The function f(t)=|98.6_x| Represent a person

I need the answer to number 2

gladu [14]

Subtract 10 from both sides :
-3y + 10 - 10 = -14 -10
3y = -24
divide both sides with 3 :
y = -8

8 0

4 years ago

Which of the following is equivalent to this expression?

qaws [65]

Answer:

6^8

Step-by-step explanation:

(6^2)^3*6^2

= 6^(2*3)*6^2

= 6^6*6^2

= 6^(6+2)

= 6^8

8 0

3 years ago

Grace and Zoe like to jump rope grades jump 66 times a minute Zoe jumps 56 times a minute they both Count Their jumps today and

san4es73 [151]

Answer:

Grace jumps for 9 minutes.

Zoe jumps for 12 minutes.

Step-by-step explanation:

Let the time for which Grace jumps the rope = $x$ minutes

Let the time for which Zoe jumps the rope = $y$ minutes

As per question statement:

$x+y=21 ..... (1)$

Number of times Grace jumps in one minute = 66

Number of times Zoe jumps in one minute = 56

Number of times Grace jumps in $x$ minutes = 66

Number of times Zoe jumps in $y$ minutes = 56

As per question statement:

$66x+56y=1266 ...... (2)$

Multiplying equation (1) with 66 and subtracting equation (2) from it:

$10y = 1386 - 1266\\\Rightarrow 10y=120\\\Rightarrow y =\bold{12\ min}$

By equation (1):

$x=21-12=\bold{9\ min}$

Therefore, the answer is:

Grace jumps for 9 minutes.

Zoe jumps for 12 minutes.

5 0

3 years ago

I get confused a lot with the "or" and "and". Please help, I have a test tomorrow!!

kenny6666 [7]

"and" is where they intersect

"or" is adding both circles (only add the intersection once)

Start at the end of the paragraph and work backwards:

10 people like all 3 flavors so write 10 in the middle (where all 3 intersect).

*************************************************************************************************

26 like coffee and pecan so write 16 where those two circles intersect. (26 - 10 = 16, remember that 10 are already included in both circles)

15 like mint and pecan so write 5 where those two circles intersect (15 - 10 = 5, remember that 10 are already included in both circles)

22 like mint and coffee so write 12 where those two circles intersect (22 - 10 = 12, remember that 10 are already included in both circles).

************************************************************************************************

48 like pecan so write 17 in the big circle (48 - 10 - 16 - 5). The intersections contain 5, 16, and 10, so that leaves a remainder of 17.

52 like coffee so write 14 in the big circle (52 - 10 - 16 - 12). The intersections contain 12, 16, and 10, so that leaves a remainder of 14.

30 like mint so write 3 in the big circle (30 - 10 - 12 - 5). The intersections contain 5, 12, and 10, so that leaves a remainder of 3.

*********************************************************************************************

Answers:

a) 17 (big pecan circle only)

b) 60 (sum of all numbers in both mint and coffee circles)

c) 23 (the sum of every circle and intersection is 77, subtract that from 100)

d) 32 (pecan circle and intersections but not the pecan/coffee intersection)

e) 16 (intersection of pecan and coffee without the mint)

4 0

4 years ago

7.2 Given a test that is normally distributed with a mean of 100 and a standard deviation of 10, find: (a) the probability that

kompoz [17]

Answer:

a)

The probability that a single score drawn at random will be greater than 110

P( X > 110) = 0.1587

b)

The probability that a sample of 25 scores will have a mean greater than 105

P( x> 105) = 0.0062

c)

The probability that a sample of 64 scores will have a mean greater than 105

P( x⁻> 105) = 0.002

d)

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = 0.9544

Step-by-step explanation:

a)

Given mean of the Normal distribution 'μ' = 100

Given standard deviation of the Normal distribution 'σ' = 10

a)

Let 'X' be the random variable of the Normal distribution

let 'X' = 110

$Z = \frac{x-mean}{S.D} = \frac{110-100}{10} =1$

The probability that a single score drawn at random will be greater than 110

P( X > 110) = P( Z >1)

= 1 - P( Z < 1)

= 1 - ( 0.5 +A(1))

= 0.5 - A(1)

= 0.5 -0.3413

= 0.1587

b)

let 'X' = 105

$Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{25} } } = 2.5$

The probability that a single score drawn at random will be greater than 110

P( x> 105) = P( z > 2.5)

= 1 - P( Z< 2.5)

= 1 - ( 0.5 + A( 2.5))

= 0.5 - A ( 2.5)

= 0.5 - 0.4938

= 0.0062

The probability that a single score drawn at random will be greater than 105

P( x> 105) = 0.0062

c)

let 'X' = 105

$Z = \frac{x-mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{64} } } = 4$

The probability that a single score drawn at random will have a mean greater than 105

P( x> 105) = P( z > 4)

= 1 - P( Z< 4)

= 1 - ( 0.5 + A( 4))

= 0.5 - A ( 4)

= 0.5 - 0.498

= 0.002

The probability that a sample of 64 scores will have a mean greater than 105

P( x⁻> 105) = 0.002

d)

Let x₁ = 95

$Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{95-100}{\frac{10}{\sqrt{16} } } = -2$

Let x₂ = 105

$Z = \frac{x_{1} -mean}{\frac{S.D}{\sqrt{n} } } = \frac{105-100}{\frac{10}{\sqrt{16} } } = 2$

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = P( -2≤z≤2)

= P(z≤2) - P(z≤-2)

= 0.5 + A( 2) - ( 0.5 - A( -2))

= A( 2) + A(-2) (∵A(-2) =A(2)

= A( 2) + A(2)

= 2 × A(2)

= 2×0.4772

= 0.9544

The probability that the mean of a sample of 16 scores will be either less than 95 or greater than 105

P( 95 ≤ X≤ 105) = 0.9544

7 0

3 years ago