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3 years ago

What’s the mean of these numbers: 7, 14, 15, 9, 11, 14, 11, 10, and 17

Mathematics

2 answers:

harkovskaia [24]3 years ago

7 0

Answer:

Step-by-step explanation:

Mean is the average of all the numbers. Add up all the numbers then divide by how many you have.

Morgarella [4.7K]3 years ago

7 0

Answer:

The mean of the numbers is 12.

Step-by-step explanation:

Mean (Average) 12

Median 11

Range 10

Mode 14, 11, each appeared 2 times

Geometric Mean 11.602375261126

Largest 17

Smallest 7

Sum 108

Count 9

Sorted Data Set: 7, 9, 10, 11, 11, 14, 14, 15, 17

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What is24/3 as a mixed number

Artist 52 [7]

24/3 in a mixed fraction is 8

7 0

3 years ago

△ABCis reflected to form △A′B′C′. The vertices of △ABC are A(3, 1), B(1, 5), and C(6, 9). The vertices of △A′B′C′ are A′(−1,

Blababa [14]

Answer: D) reflection across y = -x

Explanation:

When we reflect over y = x, we basically swap x and y. So for instance, the point (3,1) becomes (1,3).

When reflecting over y = -x, we will do the same thing but we'll make each coordinate swap in sign from positive to negative (or vice versa). The rule for reflecting over y = -x is $(x,y) \to (-y,-x)$

So if we apply that rule to point A(3,1) then it becomes A ' (-1, -3).

Similarly, B(1,5) moves to B ' (-5, -1)

Finally, C(6,9) becomes C ' (-9, -6)

5 0

2 years ago

A simple random sample of size n equals 200 individuals who are currently employed is asked if they work at home at least once p

____ [38]

Answer: 99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week

//0.20113,0.20887[/tex]

Step-by-step explanation:

step 1:-

Given sample size n=200

of the 200 employed individuals surveyed 41 responded that they did work at home at least once per week

Population proportion of employed individuals who work at home at least once per week P = $\frac{x}{n} =\frac{41}{200} =0.205$

Q=1-P= 1-0.205 = 0.705

step 2:-

Now $\sqrt{\frac{P Q}{n} } =\sqrt{\frac{(0.205)(0.705)}{200} }$

=0.0015

step 3:-

Confidence intervals

using formula

$(P - Z_∝} \sqrt{\frac{P Q}{n},} (P + Z_∝} \sqrt{\frac{P Q}{n},$

$(0.205-2.58(0.0015),0.205+2.58(0.0015)\\0.20113,0.20887$

=0.20113,0.20887[/tex]

conclusion:-

99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week

//0.20113,0.20887[/tex]

4 0

3 years ago

Sixty seven percent of the employees in a company have managerial positions, and 58 percent of the employees in the company have

Kazeer [188]

Answer: The proportion of employees who either have MBAs or are managers are 0.58.

Step-by-step explanation:

Since we have given that

Probability of employees having managerial positions = 67%

Probability of employees having MBA degrees = 58%

Probability of managers having MBA degrees = 67%

So, using probability formulas, we get that

$P(A\cup B)=P(A)+P(B)-P(A\cap B)\\\\P(A\cup B)=0.67+0.58-0.67\\\\P(A\cup B)=0.58$

Hence, the proportion of employees who either have MBAs or are managers are 0.58.

7 0

3 years ago

A soccer team wins 65% of its matches, and 15% of its matches end in a draw. If the team is scheduled to play 20 matches, about

Finger [1]

The team plays 20 matches.

65% of the matches, the teams wins
= 20 * 0.65
= 13 matches

15% of the matches, the games ends in draw.
= 20 * 0.15
= 3 matches

The team is expected to lose in
= 20 - 13 - 3
= 4 matches

3 0

3 years ago