Ask question

Ask question

All categories

3 years ago

11

I am a three digit number. I am number between 350 and 400. All my digits are different and they are all odd. The sum of my last

two digits is 16. What number am I

2 answers:

konstantin123 [22]3 years ago

4 0

Answer:

397

Explanation: 9+7 =16 and 3 is odd

Bas_tet [7]3 years ago

3 0

Answer:

397

Step-by-step explanation:

379 or 397 with fit the criteria

You might be interested in

DONT ASK FOR BRAINLIEST

KiRa [710]

2(5k+1)

2x5x2+1 =21
K=2 | 21

2x5x3+1 =31
K=3 | 31

3 0

3 years ago

The equation is y = -5(3^x)

balandron [24]

Answer:

C

Step-by-step explanation:

Since the number outside of the parentheses is -5, the negative factor means that the line is reflected across the x-axis.

3 0

3 years ago

Read 2 more answers

work for a publishing company. The company wants to send two employees to a statistics conference. To be fair, the company deci

NISA [10]

Answer:

Step-by-step explanation:

Here is the complete question.

Dominique, Marco, Roberto , and John work for a publishing company. The company wants to send two employees to a statistics conference. To be fair, the company decides that the two individuals who get to attend will have their names drawn from a hat. This is like obtaining a simple random sample of size 2. (a) Determine the sample space of the experiment. That is, list all possible simple random samples of size n = 2. (b) What is the probability that Dominique and Marco attend the conference? (c) What is the probability that John attends the conference? (d) What is the probability that John stays home?

Since there are four employees to select the two to send from. then for us to get the sample space for the experiment we need to merge all possible two employees and represent them as set.

Let D = Dominique, M = Marco, R = Roberto and J = John

a) The sample space for the experiment is the total number of possible outcomes that we can have. It is as given below

S = 4C2 = 4!/(4-2)!2! (Selecting 2 out of 4 employees)

Total sample space = 4!/2!2!

Total sample space = 4*3*2!/2!2

Total sample space = 12/2 = 6

The sample space are S = {DM, DR, DJ, MR, MJ, RJ}

b) Probability is the ratio of number of event to the sample space.

P = n(E)/n(S)

Given n(S) = 6

n(E) is the event of Dominique and Marco attending the conference.

E = {DM}

n(E) = 1

P(D and M) = 1/6

Hence the probability that Dominique and Marco attend the conference is 1/6

c) For John to attend the conference, the event outcome will be given as;

E = {DJ, MJ, RJ}

n(E) = 3

n(S) = 6

Probability for John to attend the conference is 3/6 = 1/2

d) Probability that John stays at home = 1 - Prob (John attends the conference)

Probability that John stays at home = 1 - 1/2

Probability that John stays at home = 1/2

7 0

3 years ago

The sequence an = 8, 13, 18, 23, ... is the same as the sequence a1 = 8, an = an-1 + 5.

harkovskaia [24]

Actually, this answer would be true. Why?

The first equation is: a(sub <em>n</em>) = 8, 13, 18, 23

The second is: a(sub 1)=8 ; a(sub <em>n</em>)= a(sub <em>n</em>-1)+5
if you wish to find the second term, plug two into the equation for <em /><em>n</em>
8+5=13
to find the third, plug the second term, 13, in for <em>n.</em>
13+5=18.

Hope this helped! I know it's a bit on the late side, but at least you can get the general idea!

8 0

3 years ago

Read 2 more answers

A simple random sample of size nequals10 is obtained from a population with muequals68 and sigmaequals15. (a) What must be true

valentina_108 [34]

Answer:

(a) The distribution of the sample mean ( $\bar x$ ) is <em>N</em> (68, 4.74²).

(b) The value of $P(\bar X$ is 0.7642.

(c) The value of $P(\bar X\geq 69.1)$ is 0.3670.

Step-by-step explanation:

A random sample of size <em>n</em> = 10 is selected from a population.

Let the population be made up of the random variable <em>X</em>.

The mean and standard deviation of <em>X</em> are:

$\mu=68\\\sigma=15$

(a)

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Since the sample selected is not large, i.e. <em>n</em> = 10 < 30, for the distribution of the sample mean will be approximately normally distributed, the population from which the sample is selected must be normally distributed.

Then, the mean of the distribution of the sample mean is given by,

$\mu_{\bar x}=\mu=68$

And the standard deviation of the distribution of the sample mean is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{15}{\sqrt{10}}=4.74$

Thus, the distribution of the sample mean ( $\bar x$ ) is <em>N</em> (68, 4.74²).

(b)

Compute the value of $P(\bar X$ as follows:

$P(\bar X$

$=P(Z$

*Use a <em>z</em>-table for the probability.

Thus, the value of $P(\bar X$ is 0.7642.

(c)

Compute the value of $P(\bar X\geq 69.1)$ as follows:

Apply continuity correction as follows:

$P(\bar X\geq 69.1)=P(\bar X> 69.1+0.5)$

$=P(\bar X>69.6)$

$=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}}>\frac{69.6-68}{4.74})$

$=P(Z>0.34)\\=1-P(Z$

Thus, the value of $P(\bar X\geq 69.1)$ is 0.3670.

7 0

3 years ago