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2 years ago

If f(1) = 2 and f(n) = 5f(n − 1) then find the value of f(5). -

Mathematics

1 answer:

Juliette [100K]2 years ago

8 0

Answer:

f ' (n) = 5f

Step-by-step explanation:

1- Take the derivative

2- remove the parenthesis

3- use differentiation rules

4- Differentiate

5- Simplify

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Find the third term of (x^2 + 2)^4.

attashe74 [19]

Answer:

Step-by-step explanation:

$in~the~expansion~of~(x+a)^n\\T_{r+1}=ncr x^{n-r}a^r\\r+1=3\\r=3-1=2\\n=4\\4c2=\frac{4 \times 3}{2 \times 1} =6\\T_{2}=4(x^2)^{4-2}(2)^2\\=16(x^2)^2\\=16x^4$

7 0

3 years ago

Read 2 more answers

I’m little confused can any explain? (NO LINK PLS <3)

Goshia [24]

Answer:

Step-by-step explanation:

To find the range substitute the value of x from the domain into the equation.

x = - 2 : y = -5(- 2) + 3 = 10 + 3 = 13

x = 0 : y = - 5(0) + 3 = 0 + 3 = 3

x = 1 : y = - 5(1) + 3 = - 5 + 3 = - 2

x = 3 : y = - 5(3) + 3 = - 15 + 3 = - 12

range is { - 12, - 2, 3, 13 }

with greatest value in the range being 13

6 0

2 years ago

One True Love? A survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each pe

sertanlavr [38]

Answer:

99% confidence interval for the proportion of people who disagree with the statement is [0.667 , 0.713].

Step-by-step explanation:

We are given that a survey that asked whether people agree or disagree with the statement ‘‘There is only one true love for each person." has been conducted. The result is that 735 of the 2625 respondents agreed, 1812 disagreed, and 78 answered ‘‘don’t know."

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of people who disagree with the statement = $\frac{1812}{2625}$ = 0.69

n = sample of respondents = 2625

p = population proportion of people who disagree with statement

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

So, 99% confidence interval for the population proportion, p is ;

P(-2.58 < N(0,1) < 2.58) = 0.99 {As the critical value of z at 0.5% level

of significance are -2.58 & 2.58}

P(-2.58 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.58) = 0.99

P( $-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p = [ $\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.69-2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } }$ , $0.69+2.58 \times {\sqrt{\frac{0.69(1-0.69)}{2625} } }$ ]