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3 years ago

Use the Distributive property to write each expression as an equivalent algebraic expression.

Mathematics

1 answer:

Kryger [21]3 years ago

6 0

Answer:

Step-by-step explanation:

we will use distributive property to write an equivalent expression:

7(-2x)+7(6)

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How do i solve this

sesenic [268]

I worked this out, and the answer is B.) 14x+6
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3 years ago

Read 2 more answers

If you invest $823 at an interest rate of 3 percent,how much money will you have in total afer nine years?

valentinak56 [21]

Answer:

$1,045.21

Step-by-step explanation:

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3 years ago

Given a slope and a point write a perpendicular equation in standard form.

jeyben [28]

Hi there! The answer would be:

y=-2/5x+2.

Explanation:
We can start by plotting the given point (-5,3). From there, we can use the slope to place other points to find the y intercept. Using the slope of the graph, we can find that the line segment hits (2,0). Therefore, how answer would be: y=-2/5x+2

Hope this helps!!! :DD
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3 years ago

Solve 4+2h < -3 please help me and please show work

anastassius [24]

4 + 2h < -3
2h < -7
h < -3.5

x ∈ (-∞, -3.5)

7 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