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3 years ago

5. David Read's the problem:

Mathematics

1 answer:

vladimir1956 [14]3 years ago

7 0

Answer:

No, David's answer does not seem reasonable.

Let x= original purchase amount

25x = $10.25, so solve for x by dividing both sides by .25, and you get x = $41.

If the shirt was $18, the shorts must have been $41-$18 = $23 NOT $41.

Step-by-step explanation:

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Using the following triangle, what is the cosine of angle B?

stealth61 [152]

Hello!

To find cosine, use the formula cos = adjacent / hypotenuse.

According to angle B, adjacent of angle B is side A, and the hypotenuse is side c because the hypotenuse is always opposite the right angle.

Therefore, the cosine of angle B is a/c.

3 0

3 years ago

What is the y intercept of the line -4x + 6y = -12?

snow_lady [41]

-4x + 6y = -12
-4(0) + 6y = -12
   0 + 6y = 12
   <u>6y</u> = <u>12</u>
      6 6
   y = 2

The y-intercept of the equation is 2.

6 0

3 years ago

Scottie is standing next to a lighthouse on the edge of a cliff overlooking the ocean. The cliff is 30 meters above

Anit [1.1K]

Scottie = 30

Top of the lighthouse = 45 (30+15)

The boat = 0

The fish = -10

10 + 30 = 40

5 0

3 years ago

Read 2 more answers

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

The oldest child in a family of four children is twice as old as the yougest. The two middle children are 12 and 15 years old. I

vesna_86 [32]

Answer:

8 years old

Step-by-step explanation:

oldest= 16

8+12+15+16= 51

51/4 = 12.73 ( average)

3 0

3 years ago