Which of these statements is true?

with Judy jogging north and Jackie jogging west (on straight roads). When Jackie is 1 mile farther fro m the intersection than J

dolphi86 [110]

Judy's distance from the intersection: d

When Jackie is 1 mile farther from the intersection than Judy:
Jakie's distance from the intersection: d+1

The distance between them is 2 miles more than Judy’s distance from the intersection: d+2

How far is Jackie from the intersection?
d+1=?

Using the Pythagoras Theorem:
c^2=a^2+b^2; a=d, b=d+1, c=d+2
(d+2)^2=(d)^2+(d+1)^2
(d)^2+2(d)(2)+(2)^2=d^2+(d)^2+2(d)(1)+(1)^2
d^2+4d+4=d^2+d^2+2d+1
d^2+4d+4=2d^2+2d+1
d^2+4d+4-d^2-4d-4=2d^2+2d+1-d^2-4d-4
0=d^2-2d-3
d^2-2d-3=0
Factoring:
(d-3)(d+1)=0

d+1=0→d+1-1=0-1→d=-1<0 (negative): No possible. The distance can't be negative.

d-3=0→d-3+3=0+3→d=3 miles>0 (positive): Ok

Jakie's distance from the intersection=d+1=3+1→d+1=4 miles

Answer: Jackie is 4 miles far from the intersection

Please, see the attached file.
Thanks

6 0

3 years ago

What is 15% of 90? A. 14.4 B. 13.5 C. 60.0 D. 16.7

timofeeve [1]

B.

It's 13.5

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5 0

2 years ago

Read 2 more answers

Determine the domain on which the following function is decreasing.

dexar [7]

The way I remember domain(the distance where x is shown) and range(the distance where y is shown) is that it lines up with an ordered pair. It sounds wrong to say range and domain, just like it sounds wrong to say y,x.

the answer for this would be [0,11]

7 0

3 years ago

A rectangle has a length of x+5 cm and a width of x cm. The perimeter of the rectangle is 50 cm. What is the value of x?

larisa86 [58]

Answer:x=10

Step-by-step explanation:

First let's create an equation:

2(x+5)+2x=50

2x+10+2x=50

4x+10=50

4x=40

x=10

4 0

2 years ago

Read 2 more answers

Ten percent of the engines manufactured on an assembly line are defective.

antoniya [11.8K]

Answer:

a) the probability that the first non-defective engine will be found on the second trial is 0.09

b) probability that the third non-defective engine will be found on the fifth trial is 0.00486

c) the Mean is 1.1111 and Variance is 0.1235

d) the Mean is 3.3333and Variance is 0.3704

Step-by-step explanation:

Firstly;

Let x be the number of trail on which the rth defective occurs

Also let the probability that an occurrence of a defective be p = 0.10

Here, x follows negative binomial distribution with parameters r and p

The probability mass function of X is as follows:

P(X = x) = [ x -1 p^r ( 1 - p)^(x-r) ; x = r, r + 1, r + 2, ...

r - 1 ]

= [ x -1 (0.10)^r ( 1 - 0.10)^(x-r)

r - 1 ]

= [ x -1 (0.10)^r ( 0.9)^(x-r) ..........n 1..... let this be equatio

r - 1 ]

This represents the probability that the rth success occurs on the xth trail.

a)

probability that the first non-defective engine will be found on the second trial?

Substitute r = 1 and x = 2 in equation 1

P(X = 2) = [ 2 - 1 (0.10)¹ ( 0.90 )²⁻¹

1 - 1 ]

= (0.10) × (0.90)

= 0.09

Therefore, the probability that the first non-defective engine will be found on the second trial is 0.09

b)

probability that the third non-defective engine will be found on the fifth trial?

So we substitute r = 3 and x = 5 in equation 1

P(X = 5) = [ 5 - 1 (0.10)³ ( 0.90)⁵⁻³

3 - 1 ]

= [ 4 (0.10)³ ( 0.9)²

2 ]

= 0.00486

Therefore, probability that the third non-defective engine will be found on the fifth trial is 0.00486

Now the formula for the mean of the negative binomial distribution is as follows:

Mean u = r / p ------- let this be equation 2

The formula for the variance of the negative binomial distribution also is as follows:

Variance α² = rq / p² ---------- let this be equation 3

so

c)

the mean and variance of the number of trials on which the first non-defective engine is found.

First, let the probability that non-defective engine found be p = 0.90

And q = (1 - p) = 1 - 0.90 = 0.10

Now we substitute r = 1, p = 0.90 and q = 0.10 in equation 2 & 3simultaneously,

the mean and variances are as follows;

Mean = r/p = 1/0.90 = 1.1111

Variance = rq/p² = (1)(0.10) / (0.90)² = 0.1235

Therefore the Mean is 1.1111 and Variance is 0.1235

d)

the mean and variance of the number of trials on which the third non-defective engine is found

Let the probability that non-defective engine found be p = 0.90

And q = (1 - p) = 1 - 0.90 = 0.10

Now we substitute r = 3, p = 0.90 and q = 0.10 in equation (2) and (3) simultaneously,

the mean and variances are;

Mean = r/p = 3/0.90 = 3.3333

Variance = rq/p² = (3)(0.10) / (0.90)² = 0.3704

Therefore the Mean is 3.3333 and Variance is 0.3704

5 0

3 years ago