Pls awnser asap!!!!!!!!!!!!!!!!!!!!!!!!!!!

Can you help me with this please. Please show your work. Thx in advance

pentagon [3]

That is correct I think it is

4 0

3 years ago

Read 2 more answers

35

jeka94

Answer C

steps

deduct 10 from 26

26-10 which will give us 16

5 0

3 years ago

william is thinking of two numbers both numbers are square numbers greater than 1 the sum of numbers is 100 write down two numbe

zheka24 [161]

Answer:

Step-by-step explanation:

At first,

Let start writing the squares of number 0 to 10,

0²=0

1²=1

2²=4

3²=9

4²=16

5²=25

6²=36

7²=49

8²=64

9²=81

10²=100

Now,

According to your question,

The two numbers should be square numbers
They should be greater than 1.
Their sum should be 100.

Hence, your given conditions matches with the numbers 64 and 36.

Therefore, the numbers are 64 and 36.

64 and 36 are greater than 1.
64 is a square of 8 and 36 is the square of 6.
64+36=100

3 0

3 years ago

Which measure can be used to measure the variation of this data? Find this measure.

andriy [413]

What is the measure?

5 0

3 years ago

The number of people arriving for treatment at an emergency room can be modeled by aPoisson process with a rate parameter of 5/h

saveliy_v [14]

Answer:

(a) The probability of having exactly four arrivals during a particular hour is 0.1754.

(b) The probability that at least 3 people arriving during a particular hour is 0.7350.

(c) The expected arrivals in a 45 minute period (0.75 hours) is 3.75 arrivals.

Step-by-step explanation:

(a) If the arrivals can be modeled by a Poisson process, with λ = 5/hr, the probability of having exactly four arrivals during a particular hour is:

$P(X=4)=\frac{\lambda^{X}*e^{-\lambda} }{X!} =\frac{5^{4}*e^{-5} }{4!}=\frac{625*0.006737947}{24} =\frac{4.211}{24}=0.1754$

The probability of having exactly four arrivals during a particular hour is 0.1754.

(b) The probability that at least 3 people arriving during a particular hour can be written as

$P(X>3)=1-P(X\leq3)=1- (P(0)+P(1)+P(2)+P(3))$

Using

$P(X)=\frac{\lambda^{X}*e^{-\lambda} }{X!}$

We get

$P(X>3)=1-P(X\leq3)=1- (P(0)+P(1)+P(2)+P(3))\\P(X>3)=1-(0.0067+ 0.0337+ 0.0842 + 0.1404 )\\P(X>3)=1-0.2650=0.7350\\$

The probability that at least 3 people arriving during a particular hour is 0.7350.

(c) The expected arrivals in a 45 minute period (0.75 hours) is

$EV=\lambda*t=5*0.75=3.75$

8 0

3 years ago