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klasskru [66]
2 years ago
7

Jacob wants to know the favorite sport among 6th grade students. Which types of survey below would give Jacob valid results?

Mathematics
2 answers:
agasfer [191]2 years ago
8 0

Answer:

C

Step-by-step explanation:

A is no use for 6th grade students

And B and D are bias because they are surveyed at sport places

Anastasy [175]2 years ago
4 0
Answers are A, D
Because they will have people that don’t like sports and do like sports where as if you survey them in a sports shop there gonna like sports otherwise they wouldn’t be there
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Question 3 (2 marks)
Sav [38]

Answer:

$3385 per week

Step-by-step explanation:

so, he gets $370 plus 4.5% of what he is selling.

$67000 = 100% of selling

1% = 100%/100 = 67000/100 = $670

4.5% = 1% × 4.5 = 670 × 4.5 = $3015

so, he gets $370 + $3015 = $3385 per week

7 0
2 years ago
The amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and standard deviation 13 mL. Supp
andreyandreev [35.5K]

Answer:

(a) X ~ N(\mu=63, \sigma^{2} = 13^{2}).

    \bar X ~ N(\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

Step-by-step explanation:

We are given that the amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and a standard deviation of 13 mL.

Suppose that 43 randomly selected people are observed pouring syrup on their pancakes.

(a) Let X = <u><em>amount of syrup that people put on their pancakes</em></u>

The z-score probability distribution for the normal distribution is given by;

                      Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = mean amount of syrup = 63 mL

            \sigma = standard deviation = 13 mL

So, the distribution of X ~ N(\mu=63, \sigma^{2} = 13^{2}).

Let \bar X = <u><em>sample mean amount of syrup that people put on their pancakes</em></u>

The z-score probability distribution for the sample mean is given by;

                      Z  =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \mu = mean amount of syrup = 63 mL

            \sigma = standard deviation = 13 mL

            n = sample of people = 43

So, the distribution of \bar X ~ N(\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < X < 62.8 mL)

   P(61.4 mL < X < 62.8 mL) = P(X < 62.8 mL) - P(X \leq 61.4 mL)

  P(X < 62.8 mL) = P( \frac{X-\mu}{\sigma} < \frac{62.8-63}{13} ) = P(Z < -0.02) = 1 - P(Z \leq 0.02)

                                                           = 1 - 0.50798 = 0.49202

  P(X \leq 61.4 mL) = P( \frac{X-\mu}{\sigma} \leq \frac{61.4-63}{13} ) = P(Z \leq -0.12) = 1 - P(Z < 0.12)

                                                           = 1 - 0.54776 = 0.45224

Therefore, P(61.4 mL < X < 62.8 mL) = 0.49202 - 0.45224 = 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < \bar X < 62.8 mL)

   P(61.4 mL < \bar X < 62.8 mL) = P(\bar X < 62.8 mL) - P(\bar X \leq 61.4 mL)

  P(\bar X < 62.8 mL) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{62.8-63}{\frac{13}{\sqrt{43} } } ) = P(Z < -0.10) = 1 - P(Z \leq 0.10)

                                                           = 1 - 0.53983 = 0.46017

  P(\bar X \leq 61.4 mL) = P( \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } } \leq \frac{61.4-63}{\frac{13}{\sqrt{43} } } ) = P(Z \leq -0.81) = 1 - P(Z < 0.81)

                                                           = 1 - 0.79103 = 0.20897

Therefore, P(61.4 mL < X < 62.8 mL) = 0.46017 - 0.20897 = 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

4 0
3 years ago
The chart shows the low temperature each day in International Falls, Minnesota, during the month of February 2009. What was the
Nesterboy [21]
I think you have to list all of the temperatures in order from least to greatest during the first week (-31, -29, -9, -8, 3, 4, 6) in which case the median would be -8.

I hope this helps you and if you know the answer is wrong please let me know in the comments and I will take another look at the question to see if I can figure out the correct answer.
6 0
3 years ago
Read 2 more answers
Find the inverse of g(x)=1/3x - 7 and then find its domain and range if you can show steps that would be gladly appreciated than
Schach [20]

Answer:

g-1(x) = 3(x + 7).

Domain is All Real values of x.

Range is All Real values of the function.

Step-by-step explanation:

Let y = 1/3 x - 7

1/3 x = y + 7

Multiply both sides by 3:

x = 3(y + 7)

So the inverse of g(x) = g-1(x) = 3(x + 7).

Domain is All Real values of x.

Range is All Real values of the function.

6 0
3 years ago
I need help with this can you?
Reptile [31]

Answer:

A and  D

Step-by-step explanation:

pls make me brainliest

7 0
3 years ago
Read 2 more answers
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