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Sati [7]
3 years ago
6

Can someone help me with this please?

Mathematics
1 answer:
Simora [160]3 years ago
5 0
The answer would be C
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Find the remainder when 1 + 2 + 2^2 + 2^3 + ... + 2^100 is divided by 7.<br><br> Thanks in advance!
Firlakuza [10]

Answer:

3.

Step-by-step explanation:

This is a geometric series so the sum is:

a1 * r^n - 1 / (r - 1)

= 1 * (2^101 -1) / (2-1)

= 2^101 - 1.

Find the remainder when 2^101 is divided by 7:

Note that 101 = 14*7 + 3 so

2^101 = 2^(7*14 + 3) =  2^3  * (2^14)^7 = 8 * (2^14)^7.

By Fermat's Little Theorem  (2^14) ^ 7 = 2^14 mod 7 = 4^7 mod 7.

So 2^101 mod 7 = (8 * 4^7) mod 7

                           = (8 * 4) mod 7

                            = 32 mod 7

                            = 4 = the remainder when 2^101 is divided by 7.

So the remainder when 2^101- 1 is divided by 7 is 4 - 1 = 3..

4 0
3 years ago
Can you guys help me please
Nat2105 [25]

Answer:

a = 42

Step-by-step explanation:

2a + 16 + 80 = 2a + 96

2a + 96 = 180

2a = 84

a = 42

7 0
3 years ago
In a random sample 765 adults in the United States, 322 say they could not cover a $400 unexpected expense without borrowing mon
Volgvan

Answer:

a) The population under consideration in the data set is "adults in the United States".

b) The parameter being estimated is the proportion of adults in the U.S. that could not cover a $400 unexpected expense without borrowing money or going into debt.

c) The point estimate for the parameter of the population is p-hat.

d) The standard deviation of the sampling distribution

e)

\hat p=322/765=0.421\\\\s=\sqrt{\frac{\hat p(1-\hat p)}{n} } =\sqrt{\frac{0.421\cdot 0.579}{765} } =\sqrt{0.00032} =0.018

f) Yes, 50% is a value not probable if the estimations are right.

g) No, it does not change.

Step-by-step explanation:

f) We can calculate the plausability of the value 50% with our estimations.

The P-value of a proportion of 50% is very low, so the value should surprise.

z=(x-\hat p)/s=(0.5-0.421)/0.018=0.079/0.018=4.38\\\\P(z>4.38)=0

g) We re-calculate with p=0.4

z=(x-\hat p)/s=(0.5-0.4)/0.018=0.100/0.018=5.55\\\\P(z>5.55)=0

6 0
3 years ago
I need help on 31 and 33
irakobra [83]

31.

8 + x/3 = -2

Substract both sides by 8:

x/2 = -10

Multiply:

x = -20

33.

6 * (x + 15) = -42

We can divide both sides by 6:

x + 15 = -7

Substract both sides by 15:

x = -22

7 0
3 years ago
Read 2 more answers
BRAINLIESTTT ASAP! PLEASE HELP ME :)
yan [13]

Answer:

<em>P=1620</em>

<em>Third option</em>

Step-by-step explanation:

<u>Horizontal Asymptotes</u>

A given function is said to have a horizontal asymptote in y=a, if:  

\displaystyle \lim _{x\rightarrow -\infty }f(x)=a

Or,

\displaystyle \lim _{x\rightarrow +\infty }f(x)=a

For the given function, the population of the species of bird is given by :

\displaystyle p(t)=\frac{1620}{1+1.15e^{-0.042t}}

Where t is the time in years. To find the horizontal asymptote, we should compute both limits to check if they exist.  

\displaystyle \lim _{x\rightarrow +\infty }\frac{1620}{1+1.15e^{-0.042t}}=\frac{1620}{1+0}=1620

When t tends to plus infinity, P tends to 1620 .

The second asymptote is computed by:

\displaystyle \lim _{x\rightarrow -\infty }\frac{1620}{1+1.15e^{-0.042t}}=\frac{1620}{1+\infty}=0

When t tends to minus infinity, P tends to zero. Since the domain of P is t\geq 0, this asymptote is not valid, thus our only asymptote is

\boxed{P=1620}

6 0
3 years ago
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