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2 years ago

14

Pls give the answer as soon as possible!

2 answers:

Arlecino [84]2 years ago

4 0

Answer:

option 2 is the correct answer of this question

Mice21 [21]2 years ago

3 0

Answer:

1 1/6 plz give brainiest when someone else answers thx

Step-by-step explanation:

You might be interested in

A school administrator bought 6 desk sets. Each set consists of one table and one chair. He paid $48.50 for each chair. If he pa

zloy xaker [14]

Answer: The cost of each table was $62

Step-by-step explanation:

$48.50 x 6

= $291 this is the total cost of the chairs

$663 - $291

= $372 this is the overall cost of the tables

$372 ÷ 6

= $62 this is the cost of each table

5 0

2 years ago

Marking brainliest <br><br> Can someone give me an example on why animal testing is bad?

DENIUS [597]

Wdym by animal testing? Like doing experiments on rats as an example?

4 0

3 years ago

Can you guys please help me with this? :)

xxMikexx [17]

Answer:

bob

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

HELP! PLZ I NEED THIS ASAP

harina [27]

Answer:

$2x^{4}+x^{3}-x+2$

Step-by-step explanation:

To find which one is prime, let's try to factor them all. We can use the factoring by grouping method.

$x^{3}+3x^{2}-2x-6$

$x^{3}+3x^{2}$ and $-2x-6$

$x^{2}(x+3)$ and $-2(x+3)$

So this one is not prime, since you can still factor it.

$x^{3}+2x^{2}-3x-6$

$x^{3}+2x^{2}$ and $-3x-6$

$x^{2}(x-2)$ and $3(x-2)$

So this one is not prime, since you can still factor it.

$4x^{4}+4x^{2}-2x-2$

$4x^{4}+4x^{2}$ and $-2x-2$

$4x^{3}(x+1)$ and $-2(x+1)$

So this one is not prime, since you can still factor it.

$2x^{4}+x^{3}-x+2$

$2x^{4}+x^{3}$ and $-x+2$

$x^{3}(2x+1)$ and -x+2 cannot be further factored.

Therefore, $2x^{4}+x^{3}-x+2$ is a prime.

6 0

3 years ago

It is believed that 3% of people actually have this predisposition. The genetic test is 99% accurate if a person actually has th

fiasKO [112]

Answer:

Required Probability = 0.605

Step-by-step explanation:

Let Probability of people actually having predisposition, P(PD) = 0.03

Probability of people not having predisposition, P(PD') = 1 - 0.03 = 0.97

Let PR = event that result are positive

Probability that the test is positive when a person actually has the predisposition, P(PR/PD) = 0.99

Probability that the test is positive when a person actually does not have the predisposition, P(PR/PD') = 1 - 0.98 = 0.02

So, probability that a randomly selected person who tests positive for the predisposition by the test actually has the predisposition = P(PD/PR)

Using Bayes' Theorem to calculate above probability;

P(PD/PR) = $\frac{P(PD)*P(PR/PD)}{P(PD)*P(PR/PD)+P(PD')*P(PR/PD')}$

= $\frac{0.03*0.99}{0.03*0.99+0.97*0.02}$ = $\frac{0.0297}{0.0491}$ = 0.605 .

7 0

3 years ago