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Aleks [24]
3 years ago
7

What is the solution of the inequality shown below? w + 7 ≥ 2​

Mathematics
2 answers:
gregori [183]3 years ago
6 0

Answer:

w \geq -5

Step-by-step explanation:

Hope this helps! :D

Romashka [77]3 years ago
4 0

Answer:

W  ≥ -5

Step-by-step explanation:

Subtract 7 from both sides to get W by itself. 2-7=-5

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A container has 3 apple juice boxes, 4 grape juice boxes and 5 orange juice boxes. Alex picks one juice box, drinks it and then
Eduardwww [97]

Hey!

Grape over total: \frac{4}{12}

After he drinks 1, there will be 1 less. That means it will be over 11

Orange over total: \frac{5}{11}

Now multiply the fractions

\frac{4}{12} \times \frac{5}{11} = \frac{20}{132} \rightarrow \frac{10}{66} \rightarrow \frac{5}{33}

That means the probability is: \frac{5}{33}

Good luck and hope this helps! :)

4 0
4 years ago
which method for comparing fractions were you unaware of before reading? use it to compare 2/3 and 3/4​
wolverine [178]

Good morning

______

Answer:

2/3 < 3/4​

___________________

Step-by-step explanation:

2/3 = (2*4)/(3*4)=8/12

3/4 = (3*3)/(4*3)=9/12

Since 9/12>8/12 then 3/4>2/3

:)

6 0
3 years ago
What is the growth factor of the following example? Assume time is measured in the units given.
My name is Ann [436]

Answer:

B. 0.20 per century

Step-by-step explanation:

I calculated it logically

8 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
Plz, Solveeeeeeeeeeeeeeeeeeee! I will give you all my points
Blababa [14]
1/2bh
1/2b4 = 18
b4= 36
b= 9 yards
7 0
3 years ago
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