Write the standard form of the line that passes through the given points.

Please help again, sorry for asking…

prohojiy [21]

Answer:

BC=6x+4= 6x4+4 = 28 is the correct answer

AB+BC=AC

4x+8+6x+4=52

10x+12=52

10x=40

X=4

4 0

2 years ago

Read 2 more answers

Aplumber charges a $75 flat fee for jobs lasting up to an hour and $30 for each hour of labor after the first hour. Which

tamaranim1 [39]

<h3>Answer:- </h3>

$\longrightarrow \sf 75+ 30(h-1)$

<h3>Solution:-</h3>

Given that :-

job last for h hours
Cost of job lasting more than 1 hour = $30
Cost for the first hour = $75

$\leadsto$ Remaining hours for which the job will last after first hour= (h-1)

Therefore,

Cost for subsequent hours = 30×(h-1)

= 30(h-1)

Total Cost for job = Cost for 1st hour + Cost for subsequent hours

$\quad\quad\quad\quad\quad\quad\quad\quad\boxed{\sf{= 75+30(h-1)}}$

7 0

2 years ago

P(a)=0.50 p(b)=0.30 and p(a and b)=0.15 what is p(a or b)

Softa [21]

Answer:

Option C is correct

P(A or B) = 0.65

Step-by-step explanation:

Given:

P(A) =0.5

P(B)=0.30

P(A and B) =0.15

( The probability of the happening of both independent events will be there product) P( A and B ) =P(A).P(B)

Solution:

To find the probability of the Happening of event A either event B we will use the following formula

P(A or B) = P(A) + P(B)-P(A and B)

= 0.5 + 0.3 - 0.15

=0.65

6 0

3 years ago

Read 2 more answers

A national grocery store chain wants to test the difference in the average weight of turkeys sold in Detroit and the average wei

Arte-miy333 [17]

Answer:

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

The null hypothesis is accepted .

Assume the population variances are approximately the same

Step-by-step explanation:

Explanation:-

Given data a random sample of 20 turkeys sold at the chain's stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds

The first sample size 'n₁'= 20

mean of the first sample 'x₁⁻'= 17.53 pounds

standard deviation of first sample S₁ = 3.2 pounds

Given data a random sample of 24 turkeys sold at the chain's stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds

The second sample size n₂ = 24

mean of the second sample "x₂⁻"= 14.89 pounds

standard deviation of second sample S₂ = 2.7 pounds

Null hypothesis:-H₀: The Population Variance are approximately same

Alternatively hypothesis: H₁:The Population Variance are approximately same

Level of significance ∝ =0.05

Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42

Test statistic :-

$t = \frac{x^{-} _{1} - x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} } }$

where $S^{2} = \frac{n_{1} S_{1} ^{2}+n_{2}S_{2} ^{2} }{n_{1} +n_{2} -2}$

$S^{2} = \frac{20X(3.2)^2+24X(2.7)^2}{20+24-2}$

substitute values and we get S² = 40.988

$t= \frac{17.53-14.89 }{\sqrt{40.988(\frac{1}{20} }+\frac{1}{24} )}$

t = 1.3622

Calculated value t = 1.3622

Tabulated value 't' = 2.081

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

Conclusion:-

The null hypothesis is accepted

Assume the population variances are approximately the same.

6 0

2 years ago

I NEED THE ANSWER ITS DUE SOON !!! Points will be rewarded

DENIUS [597]

Answer:

Jaden added instead of subtracting 23 and 5

Step-by-step explanation:

3 0

2 years ago