Q7:<br> Find the greatest common factor of 24,36 and 48

Which of the following square roots would not be between 4 and 5?

expeople1 [14]

Answer:

√10.

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7 0

3 years ago

Read 2 more answers

When a negative base is raised to the power of an even number, the answer is always

serg [7]

Answer:

When a negative base is raised to the power of an even number, the answer is always positive.

Step-by-step explanation:

5 0

4 years ago

A given field mouse population satisfies the differential equation dp/dt=.4p-450 where p is the number of mice and t is the time

Flura [38]

Answer:

a) 7.78 months

b) 1116

Step-by-step explanation:

Given -

$\frac{dP}{dt} = 0.4p-450$

Integrating the above equation with respect to time, we get -

$\int\ \frac{dP}{0.4p-450} = \int\ dt\\$

let us define new variable x

$x = 0.4p - 450 \\dx = 0.4 dy$

substituting these values in above integral equation, we get -

$\frac{1}{0.4} \int\ \frac{dx}{x} = \int\ dt\\ln x = 0.4 t + C\\x = ce^{0.4t}$

$P (t) = \frac{C}{0.4} e^{0.4t} +1125\\$

at t = 0, P (0) = 1075, using this condition, we get -

$\frac{c}{0.4} = -50\\P(t) = 50 e^{0.4t} +1125 \\t = 7.78\\P(0) = 1125 - \frac{1125}{e^{4.8}} = 1116$

7 0

3 years ago

Someone please help me ASAP

ivann1987 [24]

Answer:

the percentage share for BBC2 remained almost the same at about 11 % each year

if you look at the chart the BBC2 almost remains stable between 10 and 12 %

1980 ( between 39 and 51)

1985 ( between 37 and 49 ) and so on

( these numbers are not exactly the same , it is about or approximately)

7 0

4 years ago

Cystic fibrosis (CF) is a hereditary lung disorder that often results in death. It can be inherited only if both parents are car

OLEGan [10]

Answer:

0.9959 = 99.59% probability that a randomly chosen person of European ancestry does not carry an abnormal CF gene given that he /she tested negative.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Tested negative

Event B: Does not carry an abdornal CF gene.

Probability of a negative test:

10% of 1/25 = 0.04 = 4%

100% of 100 - 4 = 96%. So

$P(A) = 0.1*0.04 + 1*0.96 = 0.964$

Negative test and not carrying an abdornal gene:

100% of 96%. So

$P(A \cap B) = 1*0.96 = 0.96$

Compute the probability that a randomly chosen person of European ancestry does not carry an abnormal CF gene given that he /she tested negative.

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.96}{0.964} = 0.9959$

0.9959 = 99.59% probability that a randomly chosen person of European ancestry does not carry an abnormal CF gene given that he /she tested negative.

8 0

3 years ago

Q7: Find the greatest common factor of 24,36 and 48