What is the common ratio for the geometric sequence?<br><br> 35/2, 7, 14/5, 28/25, ..…

I need help please and thank you

Y_Kistochka [10]

the surface area is 72 sq cm

hooe this helped

5 0

3 years ago

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The length of a side of an equilateral triangle is x-45

Ilia_Sergeevich [38]

Answer:

that means all the other sides are x-45 and each of the angles are 60 degrees

(i dont know what your asking for)

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.

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

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

Add: x^5 - 4x^4 + 7x^3 + 8, 9x^3+ 7x^2 - 10, and -2x^5 + 7x^4 - 3x + 8.

Nastasia [14]

Answer:

$\large\boxed{-x^5+3x^4+16x^3+7x^2-3x+6}$

Step-by-step explanation:

$(x^5 - 4x^4 + 7x^3 + 8)+(9x^3+ 7x^2 - 10)+(-2x^5 + 7x^4 - 3x + 8)\\\\=x^5 - 4x^4 + 7x^3 + 8+9x^3+ 7x^2 - 10-2x^5 + 7x^4 - 3x + 8\\\\\text{combine like terms}\\\\=(x^5-2x^5)+(-4x^4+7x^4)+(7x^3+9x^3)+7x^2-3x+(8-10+8)\\\\=-x^5+3x^4+16x^3+7x^2-3x+6$

7 0

3 years ago

Plz help!!! due tonight I need help!!!!

Pachacha [2.7K]

Answer:

14400

Step-by-step explanation: i culd be wrong tho

3 0

2 years ago

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What is the common ratio for the geometric sequence? 35/2, 7, 14/5, 28/25, ..…