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Ksivusya [100]
3 years ago
15

Help me please!..

Mathematics
1 answer:
AlexFokin [52]3 years ago
8 0

Answer:

44 times

Step-by-step explanation:

There is a 50% (2/4) chance that this could occur, as given in the question.

88*.5 = 44.

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Almonds cost $5.20 per pound Raisins cost $2.75 per pound Priya spent $11.70 of buying almonds & raisins Equation: 5.20+2.75
murzikaleks [220]

Answer:

If she bought 2lbs of almonds, she bought 0.4lbs of raisins. If she bought  1lb of almonds, she can buy up to 2.36lbs of raisins. If she bought 0.64lbs of almonds, she can buy up to 1.9lbs of almonds.

Step-by-step explanation:

6 0
4 years ago
30 points! please help
Trava [24]
Gopherus have been clocked at rates 0.13 to 0.30 mph (0.05 to 0.13 m/s
5 0
4 years ago
If x and y vary inversely and x = 8 when y = 23, what is y when x = 4?​
Lemur [1.5K]
11.5 because all you do is divide the numbers by 2
8 0
3 years ago
Read 2 more answers
If \dfrac{a}{b}=2 ​b ​ ​a ​​ =2start fraction, a, divided by, b, end fraction, equals, 2, what is the value of \dfrac{4b}{a} ​a
kakasveta [241]
So, we have two equations. The first one is a/b = 2, and the other one is to find the value of 4b/a. In order to solve this, let's find the value of b/a. As you can observe, b/a is the reciprocal of a/b. Therefore, b/a is also the reciprocal of 2. This means that b/a = 1/2. Now, substituting this to the second equation, the answer would be

4b/a = 4(1/2) = 2

The answer is 2.
3 0
3 years ago
The following table relates the weights and
MA_775_DIABLO [31]

The probability of randomly chosen tall and obese individuals is 0.30.

Firstly complete the given table and find the total of each row and column.

                 Tall   Medium   Short   Total

Obese             18       28      14  60

Normal           20       51     28         99

Underweight   12       25       9  46

Total          50        104     51 205

We have to find P(Tall | Obese).

P(Tall | Obese) =  P(Tall and Obese) / P(Obese)

Now, P(Tall and Obese) = 18 ÷ 205

And, P(Obese) = 60÷205

Substituting the above values,

P(Tall | Obese) =  18÷205/ 60÷205

P(Tall | Obese) = 18/60

P(Tall | Obese) = 0.30

Hence the probability of a tall and obese individual is 0.30.

To learn more about probability, visit: brainly.com/question/11234923

#SPJ9

8 0
1 year ago
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