Please help ?????? ???

How would you write "4/1000" In decimal form?

scZoUnD [109]

Answer:

0.004

Step-by-step explanation:

Since 4/10 is 0.4 and 4/100 is 0.04, 4/1000 must be 0.004

4 0

3 years ago

Read 2 more answers

BEST ANSWER WILL BE MARKED BRAINLIEST

ddd [48]

Answer:

4 shelves

Step-by-step explanation:

First you will have to subtract the one she keeps by her bed.

27 - 1 = 26 to be put on a shelf.

To find the least number of shelves needed, you will divided the total left by eight:

26 ÷ 8 = 3.25

She can fill 3 shelves completely and still have books left over. So she needs at least 4 shelves!

Love your profile pic btw! :)

5 0

2 years ago

Read 2 more answers

If t follows a t7 distribution, find t0 such that (a) p(|t | < t0) = .9 and (b) p(t > t0) = .05.

Thepotemich [5.8K]

Answer:

a) $P(-t_o < t_7$

Using the symmetrical property we can write this like this:

$1-2P(t_7$

We can solve for the probability like this:

$2P(t_7$

$P(t_7$

And we can find the value using the following excel code: "=T.INV(0.05,7)"

So on this case the answer would be $t_o =\pm 1.895$

b) For this case we can use the complement rule and we got:

$1-P(t_7$

We can solve for the probability and we got:

$P(t_7$

And we can use the following excel code to find the value"=T.INV(0.95;7)"

And the answer would be $t_o = 1.895$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

For this case we know that $t \sim t(df=7)$

And we want to find:

Part a

$P(|t|$

We can rewrite the expression using properties for the absolute value like this:

$P(-t_o < t_7$

Using the symmetrical property we can write this like this:

$1-2P(t_7$

We can solve for the probability like this:

$2P(t_7$

$P(t_7$

And we can find the value using the following excel code: "=T.INV(0.05,7)"

So on this case the answer would be $t_o =\pm 1.895$

Part b

$P(t>t_o) =0.05$

For this case we can use the complement rule and we got:

$1-P(t_7$

We can solve for the probability and we got:

$P(t_7$

And we can use the following excel code to find the value"=T.INV(0.95;7)"

And the answer would be $t_o = 1.895$

5 0

3 years ago

Please help me with #1 & #2! thxs

taurus [48]

the answer for number 1 is

10,962,500,000

5 0

3 years ago

In a start-up company which has 20 computers, some of the computers are infected with virus. The probability that a computer is

Alex777 [14]

Answer:

(1) The probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.

(2) The probability at least 5 computers are infected is 0.949.

Step-by-step explanation:

The probability that a computer is defective is, p = 0.40.

(1)

Let X = number of computers to be tested before the 1st defect is found.

Then the random variable $X\sim Geo(p)$ .

The probability function of a Geometric distribution for k failures before the 1st success is:

$P (X = k)=(1-p)^{k}p;\ k=0, 1, 2, 3,...$

Compute the probability that the technician tests at least 5 computers before the 1st defective computer is found as follows:

P (X ≥ 5) = 1 - P (X < 5)

= 1 - [P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) + P (X = 4)]

$=1 -[(1-0.40)^{0}0.40+(1-0.40)^{1}0.40+(1-0.40)^{2}0.40\\+(1-0.40)^{3}0.40+(1-0.40)^{4}0.40]\\=1-[0.40+0.24+0.144+0.0864+0.05184]\\=0.07776\\\approx0.078$

Thus, the probability that the technician tests at least 5 computers before the 1st defective computer is 0.078.

(2)

Let Y = number of computers infected.

The number of computers in the company is, n = 20.

Then the random variable $Y\sim Bin(20,0.40)$ .

The probability function of a binomial distribution is:

$P(Y=y)={n\choose y}p^{y}(1-p)^{n-y};\ y=0,1,2,...$

Compute the probability at least 5 computers are infected as follows:

P (Y ≥ 5) = 1 - P (Y < 5)

= 1 - [P (Y = 0) + P (Y = 1) + P (Y = 2) + P (Y = 3) + P (Y = 4)] $=1-[{20\choose 0}(0.40)^{0}(1-0.40)^{20-0}+{20\choose 1}(0.40)^{1}(1-0.40)^{20-1}\\+{20\choose 2}(0.40)^{2}(1-0.40)^{20-2}+{20\choose 3}(0.40)^{3}(1-0.40)^{20-3}\\+{20\choose 4}(0.40)^{4}(1-0.40)^{20-4}]\\=1-[0.00004+0.00049+0.00309+0.01235+0.03499]\\=1-0.05096\\=0.94904$

Thus, the probability at least 5 computers are infected is 0.949.

7 0

3 years ago