All categories

3 years ago

Given the equation –2x + 3y = 18, what are the x-intercept and y-intercepts of the graph?

2 answers:

5 0

Y-intercept happens at x=0. Plug that into the equation. you'll get the y-intercept happens at (0,6). Then, to find the x-intercept y=0. Plug that into the equation. You'ff find that the x-intercept happens at (-9,0)

Aleksandr-060686 [28]3 years ago

4 0

the answer is C or your 3rd choice. i learned that u take the x first, so i took the x and got -2x=18 and got x by its self and do the same for y and then you get your x and y intercepts.

hope this helped<3

You might be interested in

The proportion of traffic fatalities resulting from drivers with high alcohol blood levels in 1982 was approximately normally di

SSSSS [86.1K]

Answer:

a) The proportion of states would you expect to have more than 65% of their traffic fatalities from drunk driving is 0.59518.

b) The 25th percentile of this distribution is a 0.5231 proportion of deaths due to drunk driving.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the percentile of this measure.

In this problem, we have that:

Mean 0.569 and standard deviation 0.068. So $\mu = 0.569, \sigma = 0.068$

a. What proportion of states would you expect to have more than 65% of their traffic fatali- ties from drunk driving?

This is the value of X when Z has a pvalue of 0.65.

Z has a pvalue of 0.65 between $Z = 0.38$ and $Z = 0.39$ . So, we use $Z = 0.385$

$Z = \frac{X - \mu}{\sigma}$

$0.385 = \frac{X - 0.569}{0.068}$

$X - 0.569 = 0.068*0.385$

$X = 0.59518$

The proportion of states would you expect to have more than 65% of their traffic fatalities from drunk driving is 0.59518.

b. What proportion of deaths due to drunk driving would you expect to be at the 25th percentile of this distribution?

This is the value of X when Z has a pvalue of 0.25.

Z has a pvalue of 0.65 between $Z = -0.67$ and $Z = -0.68$ . So, we use $Z = -0.675$

$Z = \frac{X - \mu}{\sigma}$

$-0.675 = \frac{X - 0.569}{0.068}$

$X - 0.569 = 0.068*(-0.675)$

$X = 0.5231$

The 25th percentile of this distribution is a 0.5231 proportion of deaths due to drunk driving.

3 0

3 years ago

Consider the following work breakdown structure: Time Estimates (days) Activity Precedes Optimistic Most Likely Pessimistic Star

Doss [256]

Answer: the probability that the critical path for this project will be completed within 210 days is 0.99725

Step-by-step explanation:

Given that;

Work break down is;

Time Estimates (days)

Activity Precedes Optimistic Most Likely Pessimistic

Start A,B - - -

A C,D 44 50 56

B D 45 60 75

C E 42 45 48

D F 31 40 49

E F 27 36 39

F End 58 70 82

we know that;

Expected Time = ( Optimistic Time + ( 4 × most likely time) + pessimistic time) / 6

and

Variance = [( pessimistic time - Optimistic time) / 6 ]²

ACTIVITY EXPECTED TIME VARIANCE

A (44 + (4 × 50) + 56) / 6 = 50 ((56 - 44) / 6)² = 4

B (45 + (4 × 60) + 75) / 6 = 60 ((75 - 45) / 6)² = 25

C (42 + (4 × 45) + 48) / 6 = 45 ((48 - 42) / 6)² = 1

D (31 + (4 * 40) + 49) / 6 = 40 ((49 - 31) / 6)² = 9

E (27 + (4 * 36) + 39) / 6 = 35 ((39 - 27) / 6)² = 4

F (58 + (4 * 70) + 82) / 6 = 70 ((82 - 58) / 6)² = 16

ACTIVITY DURATION ES EF LS LF SLACK

A 50 0 50 0 50 0

B 60 0 60 30 90 30

C 45 50 95 50 95 0

D 40 60 100 90 130 30

E 35 95 130 95 130 0

F 70 130 200 130 200 0

FORWARD PASS:

ES = MAXIMUM EF OF ALL PREDECESSOR ACTIVITIES; 0 IF NO PREDECESSORS ARE PRESENT.

EF = ES + DURATION OF THE ACTIVITY.

CRITICAL PATH = PATH WITH THE LONGEST COMBINED DURATION VALUE AND 0 SLACK.

CRITICAL PATH = ACEF

DURATION OF PROJECT = 200

Now the probability

VARIANCE OF CRITICAL PATH = SIGMA(VARIANCE OF THE CRITICAL PATH) = 25

STANDARD DEVIATION OF CRITICAL PATH = SQRT(VARIANCE OF CRITICAL PATH) = 5

EXPECTED TIME = DURATION OF THE PROJECT = CRITICAL PATH = 200

DUE TIME = 210

Z = (DUE TIME - EXPECTED TIME) / STANDARD DEVIATION OF CRITICAL PATH)

Z = (210 - 200) / 5 = 2

Now form the Normal Distribution Table;

Z = 0.99725

Therefore, the probability that the critical path for this project will be completed within 210 days is 0.99725

3 0

3 years ago

Venita is factoring the expression 32ab-8b . She determines the GCF and writes the factored expression as . Which best describes

kogti [31]

Let's go through the steps of factoring that Venita should take.

1.) Find the greatest common factor (GCF). We only have two terms, so that makes it pretty easy.
32 = 1, 2, 4, 8, 16, 32
8 = 1, 2, 4, 8
The greatest common factor of 32 and 8 is 8. We can also factor out a b since that term appears in each part of the original expression. The GCF and variable should go on the outside of the parentheses.
8b( )

2.) Now let's figure out what should go in the middle of the parentheses. To do this, use the original expression and divide each term. This is written in the parentheses.

32ab ÷ 8b = 4a
8b ÷ 8b = 1

This would then result in the factored expression 8b(4a - 1). You can always check this by using the distributive property. Distribute 8b out to both expressions:
8b x 4a = 32ab
8b x 1 = 8b
32ab - 8b is the expression she started with, so your factored expression works!

Now that we went through the steps to solve the factored expression, let's check her answer. The only difference between Venita's and ours is that she has 0 as the second term while we have a 1. It seems that she had subtracted the GCF from the second term instead of dividing.

5 0

3 years ago

Melanie took out a 20-year loan for $50,000 at an APR of 3.3%, compounded

stepan [7]

Answer:

She would save $17,340.10.

Step-by-step explanation:

We will use formula : $A=P(1+r)^t$