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valentinak56 [21]
3 years ago
11

Exercise 4 Find the following sums and differences using a number line model. a. -6+3

Mathematics
1 answer:
Arada [10]3 years ago
4 0
-3 Is The Answer -6+3=-3
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A chemist wants to mix a 66% alcohol solution with eight liters of 54% alcohol solution to produce a solution that is 65% alcoho
Rzqust [24]

Answer:

We have 8 liters of 54% alcohol.

We will add "x" liters of 66% alcohol to make "8 +x" liters of 65% alcohol.

54 * 8 + 66 x =   65 (8 + x)

432 + 66x = 520 + 65x

x = 88 liters

Step-by-step explanation:

3 0
3 years ago
F(x) = -x² - 7x - 5<br>Find f(-5)​
irga5000 [103]

Answer:

5

Step-by-step explanation:

f(-5)=-(-5)^2 - 7(-5) - 5

f(-5)= -25 +35 - 5

f(-5) = -25 +30

f(-5) = 5

7 0
3 years ago
Read 2 more answers
4a+(-8), if a= -2 PLZ ANSWER ASAP
Ugo [173]

Answer:

-16

Step-by-step explanation:

4a+(-8)

4(-2)+(-8)

-8+(-8)

= - 16

6 0
3 years ago
WHO WANTS BRANLIEST?!?!?!?!?!?!?!?
denis23 [38]

Answer:

I Do

Step-by-step explanation:

5 0
3 years ago
In a clinical study, volunteers are tested for a gene that has been found to increase the risk for a disease. The probability th
MrRa [10]

Answer:

a) 0.984

b) 20 people

Step-by-step explanation:

a)

If The probability that a person carries the gene is 0.1, then in a sample of 20 people, 2 should carry the gene.

Now, we want to know how many samples there are with this property.

Since we have 20 elements where 18 are alike (do not carry the gene) and 2 are alike (carry the gene), we have to compute the number of permutations of 20 elements in which 18 are alike and 2 are alike. This number is

\frac{20!}{18!2!}=190

In this 190 20-tuples there are only 3 where the 2 carriers of the gene are in the first 3 places, namely

(1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

(1,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

(0,1,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0)

where 1=carries the gene, 0=does not carry the gene.

So there are 190-3 = 187 elements in which the first 3 elements have no 2 carriers, hence the probability that 4 or more people will have to be tested until 2 of them with the gene are detected is 187/190 = 0.984 (98.4%) rounded to three decimal places.

b)

Given that the probability that a person carries the gene is 0.1, then in a sample of 20 people, 20*0.1 = 2 should carry the gene.

4 0
3 years ago
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