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tankabanditka [31]
3 years ago
7

The nine cats in a pet store were weighed. Their weights (in pounds) are given below.

Mathematics
1 answer:
liubo4ka [24]3 years ago
8 0
Hope the shown work helps you!!

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givi [52]
Bare with me a minute, I am still trying to figure it out. Here is what I have so far.

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3 years ago
DOES ANYONE KNOW HOW TO DO THIS PROBLEM???? I AM NOT SURE HOW TO SOLVE THIS PROBLEM
Elena-2011 [213]

Answer:

1. Figure 2

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3. Strongest linear relationship in Figure 1.

Step-by-step explanation:

7 0
3 years ago
A class of 40 students has 11 students who ride the bus and 10 students who are in the band. Three of the students who ride the
Alexxandr [17]
The denominator is 40
10-3=7 (number of students in the band who do not ride the bus)
7/40=0.175
0.174*100=17.5% chance
Hope this helps!
8 0
3 years ago
10 watermelons have an average weight of 4.3 pounds and 5 pumpkins have an average weight of 3.4 pounds. What is the average wei
bagirrra123 [75]
Since the average is the sum divided by the count, the sum is the average times the count.

So the sum of the weights of the watermelons is 10(4.3)=43 pounds.

The sum of the weights of the pumpkins are 5(3.4)=17 pounds

The sum of the weight of the fruits is 43+17=60 pounds

Dividing by 15 fruits gives an average item weight of 60/15 = 4 pounds

6 0
3 years ago
Read 2 more answers
In a city school, 70% of students have blue eyes, 45% have dark hair, and 30% have blue eyes and dark hair. What is the probabil
Travka [436]

<u>Answer:</u>

The correct answer option is 43%.

<u>Step-by-step explanation:</u>

We are given that 70% of students have blue eyes, 45% have dark hair, and 30% have blue eyes and dark hair.

We are to find the probability of a student getting selected will have dark hair with blue eyes.

P(D|B) = P(D∩B) / P(B)

Substituting the given values in it to get:

P(D|B) = 0.3 / 0.7 = 0.428

Rounding it to the nearest whole percent we get:

0.428 × 100 = 42.8% ≈ 43%

4 0
3 years ago
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