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3 years ago

Which regression line properly describes the data relationship in the scatterplot?

Mathematics

2 answers:

TEA [102]3 years ago

6 0

Answer:C

Step-by-step explanation:

Bc there's 3 does below and 3 dots above

Bond [772]3 years ago

3 0

Answer:c

Step-by-step explanation:

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What is the inverse of the function f(x) = 2x + 1?

BaLLatris [955]

Answer:

$f^{-1} = \frac{x-1}{2}$

Step-by-step explanation:

$f(x) = 2x+1$

Replace it with y

$y = 2x+1$

Exchange the values of x and y

$x = 2y+1$

Solve for y

$x = 2y+1$

Subtracting 1 from both sides

$2y = x-1$

Dividing both sides by 2

$y = \frac{x-1}{2}$

Replace it by $f^{-1}$

So,

$f^{-1} = \frac{x-1}{2}$

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3 years ago

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Exponent that will equal 32768

solmaris [256]

27 I think or something close to that hope that helps

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3 years ago

what is an equation to represent for (Andre withdrew spending money from his checking account that initially had $982. What was

Paul [167]

Answer:

The Answer is w - 982 = 922 because it whatever money he took from the 982 lead to 922

Step-by-step explanation:

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2 years ago

Which is the better buy on potatoes? 3 lbs. for $.90 5 lbs. for $1.60

myrzilka [38]

Answer:

3 pounds for 90 cents.

Step-by-step explanation:

The 5 pound bag costs 32 cents a pound

The 3 pound bag is 30 cents a pound

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2 years ago

An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de

Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

$P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721$

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

$P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have:

$P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528$

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

$P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}$

Then we have

$P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453$

The sum of the probabilities is

$P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981$

8 0

3 years ago