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oksian1 [2.3K]
3 years ago
6

in a geometrical progression,the product of the 2nd and 4th term is double the the 5th terms. and the sum of the first four term

s is 80 find the G.p​
Mathematics
1 answer:
ahrayia [7]3 years ago
4 0

Answer:

solve by your self dont depend on other!

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What is 1 tenth of 3.0. as a decimal
tangare [24]
Answer=0.3

1/10 of 3.0 is 0.3

3.0*(1/10)=0.3
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I shared 3 cars with my cousins I have 5 cousins and me. I want to know how many people will be in each car
tatuchka [14]

there will be 2 people in each car

5 0
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A farmer has 150 rabbits. There are 4 times as many female rabbits as male rabbits. How many male and female rabbits are there?
enyata [817]
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7 0
4 years ago
What is the sum of the sequence 152,138,124, ... if there are 24 terms?
N76 [4]

Answer: -216

Step-by-step explanation:

To solve the exercise you must use the formula shown below:

Sn=\frac{(a_1+a_n)n}{2}

Where:

a_1=152\\a_n=a_{24}

You should find  a_{24}

The formula to find it is:

a_n=a_1+(n-1)d

Where d is the difference between two consecutive terms.

d=138-152=-14

Then:

a_{24}=152+(24-1)(14)=-170

Substitute it into the first formula. Therefore, you obtain:

S_{24}=\frac{(152-170)(24)}{2}=-216

3 0
4 years ago
Read 2 more answers
The probability of being a universal donor is 6% (O-negative-blood type). Suppose that 6 people come to a blood drive.' a) What
Tcecarenko [31]

Answer:

Mean = 0.36

SD = 0.5817

P(x=3) = 0.003588

Step-by-step explanation:

Given

Let

A = Event of being a universal donor.

So:

P(A) = 0.06

n = 6

Solving (a): Mean and Standard deviation.

The mean is:

Mean = np

Mean = 6 * 0.06

Mean = 0.36

The standard deviation is:

SD = \sqrt{np(1-p)}

SD = \sqrt{6*0.06*(1-0.06)}

SD = \sqrt{0.3384}

SD = 0.5817

Solving (b): P(x = 3)

The event is a binomial event an dthe probability is calculated as:

P(x) = ^nC_x * p^x * (1-p)^{n-x}

So, we have:

P(x=3) = ^6C_3 * 0.06^3 * (1-0.06)^{6-3}

P(x=3) = ^6C_3 * 0.06^3 * (1-0.06)^3

P(x=3) = 20 * 0.06^3 * (1-0.06)^3

P(x=3) = 0.003588

7 0
3 years ago
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