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svlad2 [7]
3 years ago
5

HELP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Aleksandr [31]3 years ago
8 0

Answer:

b

Step-by-step explanation:

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Last​ year, a person wrote 126 checks. Let the random variable x represent the number of checks he wrote in one​ day, and assume
Arisa [49]

Answer:  The mean number of checks written per​ day  =0.3452

Standard deviation=0.5875

Variance  =0.3452

Step-by-step explanation:

Given : The total number of checks wrote by person in a year = 126

Assume that the year is not a leap year.

Then  1 year = 365 days

Let the random variable x represent the number of checks he wrote in one​ day.

Then , the mean number of checks wrote by person each days id=s given by :-

\lambda=\dfrac{126}{365}\approx0.3452

Since , the distribution is Poisson distribution , then the variance must equal to the mean value i.e. \sigma^2=\lambda=0.3452

Standard deviation : \sigma=\sqrt{0.3452}=0.5875372328\approx0.5875

6 0
3 years ago
Heather finished the 12-kilometer race in 2 hours. If Heather kept her pace constant, what was her rate? A. 24 km/hr. B. 14 km/h
andriy [413]

Answer: D, 6km/hr.

Step-by-step explanation:

Heather can finish a 12-kilometer race in 2 hours, and now we have to find how many kilometers she can ride/run at in 1 hour.

How many hours can Heather run in 1 hour? To solve that, we can use the equation 12 ÷ 2.

12 ÷ 2 = 6.

Therefore, if Heather keeps her pace constant, then her rate will be 6km/hr, or D.

7 0
3 years ago
Write 235 using powers of 10
vaieri [72.5K]

Answer:

(10^2)*2, 10*3, 10/2

Step-by-step explanation:

Thats my best guess to your very vague question

8 0
2 years ago
F(x)=2x^2-3x+7 evaluate f(-2)
Vilka [71]

Answer:

Step-by-step explanation:

Find two linear functions p(x) and q(x) such that (p (f(q(x)))) (x) = x^2 for any x is a member of R?

Let  p(x)=kpx+dp  and  q(x)=kqx+dq  than

f(q(x))=−2(kqx+dq)2+3(kqx+dq)−7=−2(kqx)2−4kqx−2d2q+3kqx+3dq−7=−2(kqx)2−kqx−2d2q+3dq−7  

p(f(q(x))=−2kp(kqx)2−kpkqx−2kpd2p+3kpdq−7  

(p(f(q(x)))(x)=−2kpk2qx3−kpkqx2−x(2kpd2p−3kpdq+7)  

So you want:

−2kpk2q=0  

and

kpkq=−1  

and

2kpd2p−3kpdq+7=0  

Now I amfraid this doesn’t work as  −2kpk2q=0  that either  kp  or  kq  is zero but than their product can’t be anything but  0  not  −1 .

Answer: there are no such linear functions.

6 0
2 years ago
If $8259 principal is invested in a savings account that pays 5.25% interest compounded continuously how much money will be in t
VMariaS [17]

Answer:

A(10) = $13,961.50

Step-by-step explanation:

First, convert R as a percent to r as a decimal

r = R/100

r = 5.25/100

r = 0.0525 rate per year,

Then solve the equation for A

The formula is given as:

A = Pe^rt

P = 8259

r = 0.0525

t = 10 years.

Hence,

A = 8,259.00 × e^(0.0525×10)

A = $13,961.50

Therefore, the money that will be in the account after 10 years is $13,961.50

5 0
3 years ago
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