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3 years ago

Write the equation that describes the line in slope-intercept form: slope= 4, point (3,-2) is on the line

1 answer:

8 0

Slope intercept form is y=mx+b
m=slope
b=y intercept

slope is 4
y=4x+b
subsitute if iin (x,y) form
so x=3 and y=-2 is one solution
subsitute
-2=4 times 3+b
-2=12+b
subtract 12 from both sids
-14=b

the equation is y=4x-14

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Evaluate the expression. 2^1/2^4

tekilochka [14]

Your answer is 1^8

Hope this helps!

6 0

3 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

(x2-2x+2)+(3x2-4x-10

Citrus2011 [14]

Answer:

x^2-6x-2

Step-by-step explanation: i am 100% sure

6 0

3 years ago

Which of the following inequalities is correct?

Mice21 [21]

Answer:

Step-by-step explanation:

it is option C because a positive number will always be bigger then a negative number

3 0

2 years ago

Read 2 more answers

Suppose that 15 inches of wire costs 90 cents. At the same rate, how many inches of wire can be bought for 48 cents?

tamaranim1 [39]

This can be solved by making an equivalent ratio.
The original ratio is what we know, 15 inches of wire for 90 cents.
In a ratio of inches of wire:cents, this would be 15:90.

Now for the equivalent ratio.
We don't know the number in the inches place but we do know it for the cents place.
Let's use x to represent inches of wire.
x:48 is our new ratio, and we need to find x.

Since x:48 and 15:90 are equivalent, that means the same thing that was done to 90 to get 48 has to be done to 15 to get the value of x, since the same thing must be applied to both sides.
We can find what 90 was divided by (which is what we'll have to divide 15 by) by dividing 90 by 48.
90 / 48 = 1.875

This means 48 • 1.875 = 90 and x • 1.875 = 15.
Since we don't know x though, we can isolate it by dividing both sides by 1.875.
x • 1.875 = 15
x • 1.875 / 1.875 = x
15 / 1.875 = 8
So x is 8.

Answer:
While you can be 15 inches of wire for 90 cents, you can buy 8 inches of wire for 48 cents at the same rate.

3 0

3 years ago