Factor completely n4-1

Hello, I need help with a math question, please tell me if I am right. *Refer to the photo*

Trava [24]

As I can tell the tables are right but it would be a function since none of the x's repeat

5 0

2 years ago

How do it do this ????

V125BC [204]

Because he drove 1542.75 miles in 3 days, you would have to divide 1542.75 by 3 to figure out how many miles he drove in one day.

1542.75/3 = 514.25

Then to figure out how many miles he drove per hour, you would divide 514.25 by 8.5, which is the same as 8 1/2 hours.

514.25/8.5 = 60.5

So the answer is 60.5 mph

4 0

3 years ago

Weekly demand for milk at a small grocery store is normally distributed with a mean of 251 gallons per week and standard deviati

slavikrds [6]

Solution

Given

mean = 251

standard deviation = 33

(a)

$\begin{gathered} P(Z$

Hence, X = 300

7 0

1 year ago

Read 2 more answers

En un triángulo rectángulo A es un ángulo agudo y Sen A = 4/5 ¿Cuál será el valor de Tan A?

Nonamiya [84]

Answer:

$\displaystyle \tan A=\frac{4}{3}$

Step-by-step explanation:

<u>Funciones Trigonométricas</u>

La identidad principal en trigonometría es:

$sen^2A+cos^2A=1$

Si sabemos que A es un ángulo agudo (que mide menos de 90°), su seno y coseno son positivos.

Dado que Sen A = 4/5, calculamos el coseno:

$cos^2A=1-sen^2A$

Sustituyendo:

$\displaystyle cos^2A=1-\left(\frac{4}{5}\right)^2$

$\displaystyle cos^2A=1-\frac{16}{25}$

$\displaystyle cos^2A=\frac{25-16}{25}$

$\displaystyle cos^2A=\frac{9}{25}$

Tomando raíz cuadrada:

$\displaystyle cos\ A=\sqrt{\frac{9}{25}}=\frac{3}{5}$

La tangente se define como:

$\displaystyle \tan A=\frac{sen\ A}{cos\ A}$

Substituyendo:

$\displaystyle \tan A=\frac{\frac{4}{5}}{\frac{3}{5}}$

$\displaystyle \tan A=\frac{4}{3}$

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