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3 years ago

Find the area of the figure. Round it to the nearest hundredth

Mathematics

1 answer:

aliya0001 [1]3 years ago

4 0

Answer:

Please add the figure by clicking on the link button when posting something

Step-by-step explanation:

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Drag each label to the correct location on the image.

joja [24]

Answer:

4 x (1 x 4) = 16

Step-by-step explanation:

1 x 4 = 4

4 x 4 = 16

8 0

2 years ago

3. From the table below, find Prof. Xin expected value of lateness. (5 points) Lateness P(Lateness) On Time 4/5 1 Hour Late 1/10

wariber [46]

Answer:

The expected value of lateness $\frac{7}{20}$ hours.

Step-by-step explanation:

The probability distribution of lateness is as follows:

Lateness P (Lateness)

On Time 4/5

1 Hour Late 1/10

2 Hours Late 1/20

3 Hours Late 1/20

The formula of expected value of a random variable is:

$E(X)=\sum x\cdot P(X=x)$

Compute the expected value of lateness as follows:

$E(X)=\sum x\cdot P(X=x)$

$=(0\times \frac{4}{5})+(1\times \frac{1}{10})+(2\times \frac{1}{20})+(3\times \frac{1}{20})\\\\=0+\frac{1}{10}+\frac{1}{10}+\frac{3}{20}\\\\=\frac{2+2+3}{20}\\\\=\frac{7}{20}$

Thus, the expected value of lateness $\frac{7}{20}$ hours.

8 0

2 years ago

Jose bought a car for $32000. The car depreciates in values by 5.7% each year. Write an exponential function that represents how

lana [24]

5.7 × 5 = 28.5
Have you tried that?

3 0

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.

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