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Len [333]
3 years ago
15

Is the table right plz hlp me

Mathematics
2 answers:
Rudik [331]3 years ago
8 0
Make a better picture, I can't even see the question. 
lesya [120]3 years ago
6 0
Could you please take another picture without the flash. I could be happy to help you with your question
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Please Help!
Artist 52 [7]
X<4 hope this helped dude
5 0
4 years ago
If we have a large voluntary response sample consisting of weights of subjects who chose to respond to a survey posted on the In
Blizzard [7]

A  graph can not  overcome the deficiency of having a voluntary response sample.

<h3>What is a voluntary response sample ?</h3>

The term voluntary response sample has to do with a situation in which the researcher decides to make participation in the research a matter of the volition of the subjects. The subjects are free to respond to the survey or not.

Often time, the voluntary response sample does not accurately reflect the reality on ground due to under coverage bias. Thus, a graph can not  overcome the deficiency of having a voluntary response sample.

Learn more about  voluntary response sample:brainly.com/question/1413932

#SPJ1

3 0
2 years ago
A soccer player has a limit of 700 calories in a sandwich and chips meal. The sandwich is 464 calories and each chip is 7 calori
Blababa [14]

Answer:

33 chips

Step-by-step explanation:

c= (700 - 464) ÷ 7 = 33

She can use 33 chips or less. I less cause she shouldn't eat too much calories.  

8 0
3 years ago
Read 2 more answers
What is the slope of the line graphed on the coordinate plate? (1,-2) (0,6)
Phoenix [80]

Answer:

m = -8 (Two points slope form)

Step-by-step explanation:

y = -8x + 6

8 0
3 years ago
An airplane with room for 100 passengers has a total baggage limit of 6000 lb. Suppose that the total weight of the baggage chec
antoniya [11.8K]

Answer:

The approximate probability that the total weight of their baggage

P (\bar x >60) = 0.000

Step-by-step explanation:

Given data;

number of passenger is 100

baggage limit = 6000 lb

standard deviation = 19 lb

mean value  = 49

For 100 passengers baggage limit is 6000 lb

so, average weight for per passenger  > 60

P (\bar x >60)

As mean value is 49, therefore  P (\bar x >60) lie on right side center

z is given as

z = \frac{ \bar x \mu}{\sigma_{\bar x}}

z = \frac{60 - 49}{\frac{\sigma}{\sqrt n}}

z  = \frac{60 - 49}{\frac{19}{\sqrt {100}}} = 5.79

P(Z>5.79) = AREA to the right of 5.79

P (\bar x>60)  = P (Z>5.79)  = 1 -P (Z < 5.79)

= 1 -P (Z < 5.79)

= 1 - 1.0000

= 0.000

P (\bar x >60) = 0.000

6 0
3 years ago
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