Ask question

Ask question

All categories

3 years ago

15

My tens digit is less than 7. I am one away from being a multiple of 5. If you half me you will not get a whole number. Who am I

? A.883B.429C.133D.524E.635F.236G.712H.671

2 answers:

Hitman42 [59]3 years ago

8 0

Hope this helps, can i get brainliest please

Len [333]3 years ago

6 0

Answer:

its B.429

Step-by-step explanation:

20<70

429+1=multiple of 5

429÷2=214.5

so the answer is 429

You might be interested in

2.5% of a population are infected with a certain disease. There is a test for the disease, however the test is not completely ac

Aleks04 [339]

Answer:

$P(disease/positivetest) = 0.36116$

Step-by-step explanation:

This is a conditional probability exercise.

Let's name the events :

I : ''A person is infected''

NI : ''A person is not infected''

PT : ''The test is positive''

NT : ''The test is negative''

The conditional probability equation is :

Given two events A and B :

P(A/B) = P(A ∩ B) / P(B)

$P(B) >0$

P(A/B) is the probability of the event A given that the event B happened

P(A ∩ B) is the probability of the event (A ∩ B)

(A ∩ B) is the event where A and B happened at the same time

In the exercise :

$P(I)=0.025$

$P(NI)= 1-P(I)=1-0.025=0.975\\P(NI)=0.975$

$P(PT/I)=0.904\\P(PT/NI)=0.041$

We are looking for P(I/PT) :

P(I/PT)=P(I∩ PT)/ P(PT)

$P(PT/I)=0.904$

P(PT/I)=P(PT∩ I)/P(I)

0.904=P(PT∩ I)/0.025

P(PT∩ I)=0.904 x 0.025

P(PT∩ I) = 0.0226

P(PT/NI)=0.041

P(PT/NI)=P(PT∩ NI)/P(NI)

0.041=P(PT∩ NI)/0.975

P(PT∩ NI) = 0.041 x 0.975

P(PT∩ NI) = 0.039975

P(PT) = P(PT∩ I)+P(PT∩ NI)

P(PT)= 0.0226 + 0.039975

P(PT) = 0.062575

P(I/PT) = P(PT∩I)/P(PT)

$P(I/PT)=\frac{0.0226}{0.062575} \\P(I/PT)=0.36116$

8 0

3 years ago

Which ordered paur is a part of the equation? 3x-y=13

choli [55]

Answer:

(6,5), (3,-4) either one of those...

7 0

3 years ago

Evaluate the equation.

nordsb [41]

The answer is B) m=3 2/3

7 0

3 years ago

Read 2 more answers

You are at a stall at a fair where you have to throw a ball at a target. There are two versions of the game. In the first

Tomtit [17]

Answer:

$P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729$

And the probability of loss with the first wersion is 0.729

$P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774$

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Alternative 1

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=3, p=0.1)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

We can find the probability of loss like this P(X=0) and if we find this probability we got this:

$P(X=0)=(3C0)(0.1)^0 (1-0.1)^{3-0}=0.729$

And the probability of loss with the first wersion is 0.729

Alternative 2

Let Y the random variable of interest, on this case we now that:

$Y \sim Binom(n=5, p=0.05)$

The probability mass function for the Binomial distribution is given as:

$P(Y)=(nCy)(p)^y (1-p)^{n-y}$

Where (nCx) means combinatory and it's given by this formula:

$nCy=\frac{n!}{(n-y)! y!}$

We can find the probability of loss like this P(Y=0) and if we find this probability we got this:

$P(Y=0)=(5C0)(0.05)^0 (1-0.05)^{5-0}=0.774$

And the probability of loss with the first wersion is 0.774

As we can see the best alternative is the first version since the probability of loss is lower than the probability of loss on version 2.

4 0

3 years ago

The number of dogs at The Great North Dog walk in 2011 rounded to the nearest thousand is 23,000.Which could be the exact number

Arte-miy333 [17]

What are the choices?

4 0

3 years ago