Mathematics, 04.07.2019 19:00 dontcareanyonemo

Weight/calories per day 1000 to 1500 cal. 1500 to 2000 cal. 2000 to 2500 cal. total 120 lb. 90 80 10 180 145 lb. 35 143 25 203 165 lb. 15 27 75 117 total 140 250 110 500 based on the data in the two-way table, what is the probability that a person consumes 1,500 to 2,000 calories in a day? a. 0.22 b. 0.28 c. 0.35 d. 0.50

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Answer from: sherlock19

Option: D is the correct answer.

The probability is: 0.50

Step-by-step explanation:

Weight/Calories 1000-1500 1500-2000 2000-2500 Total

120 lb. 90 80 10 180

145 lb. 35 143 25 203

165 lb. 15 27 75 117

Total 140 250 110 500

We are asked to find the probability that a person consumes 1500-2000 calories in a day.

Total number of people who were surveyed=500

Total number of people who consume 1500-2000 calories in a day are:250

Hence, Probability is:

250/500=1/2=0.50

Answer from: Weaj01

D. 0.50

Step-by-step explanation:

Formula for probability :

$\text{Probability}=\dfrac{\text{Favorable outcomes}}{\text{Total outcomes}}$

The given two-way table :

Weight/Calories 1000-1500 1500-2000 2000-2500 Total

120 lb. 90 80 10 180

145 lb. 35 143 25 203

165 lb. 15 27 75 117

Total 140 250 110 500

To find the probability that a person consumes 1500-2000 calories in a day., we need :

a) The total number of people surveyed. (Total outcomes)

b) The number of people consume 1500-2000 calories in a day. (favorable outcomes)

From the above table,

Total people surveyed=500

Number of people consume 1500-2000 calories in a day are=250

Then , the required probability will be :-

$\text{P(person consumes 1,500 to 2,000 calories in a day)}=\dfrac{250}{500}=\dfrac{1}{2}\\\\=0.50$

Hence, the probability that a person consumes 1,500 to 2,000 calories in a day=0.50

Answer from: Quest

p> q if you calculate p is100 q is 79

What do you know to be true about the values of p and q?

Answer from: Quest

it costs $20 to interview each person for a survey. find the cost to obtain a +- 3% margin of error.

step-by-step explanation:

copy and paste, put it in google

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