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2 years ago

Evaluate.

Mathematics

2 answers:

Vadim26 [7]2 years ago

4 0

———————————-_———-ok so since (2a

liq [111]2 years ago

4 0

-1.65 should be it, can i get brainliest? :D

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Of the 17 students in a programming class, 6 own 1 computer, 3 own 2 computers, 3 own 3 computers, 2 own 4 computers, and 3 own

AlexFokin [52]

The median number of computers owned by the 17 students is 2.

<h3>What is median in statistics?</h3>

A dataset's median represents the midpoint; half of the data points are smaller and half are greater than the median.

Identifying the median: The data points should be arranged from smallest to largest. The median is the middle data point in the list if the number of data points is odd.

The total number of students are 17.

The data can be arranged in the following table

number of students number of computers

6 1

3 2

3 3

2 4

3 5

Arrange the given frequencies in the increasing order-

1 1 1 1 1 1 2 2 2 3 3 3 4 4 5 5 5

The middle term of the arrange frequencies is 17/2 = 8.5

Median = (8 term + 9 term)/2

= (2 + 2)/2

= 2

Therefore, the median number of computers owned by the 17 students is 2.

To know more about the median, here

brainly.com/question/8892632

#SPJ4

The correct question is -

Of the 17 students in a programming class, 6 own 1 computer, 3 own 2 computers, 3 own 3 computers, 2 own 4

computers, and 3 own 5 computers. What is the median number of computers owned by the 17 students?

7 0

2 years ago

Help on my recent question plz?

Mashutka [201]

Answer:

Attach it plz or I can't help

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Three cards are drawn from a standard deck of 52 cards without replacement. Find the probability that the first card is an ace,

MrRissso [65]

Answer:

$4.82\cdot 10^{-4}$

Step-by-step explanation:

In a deck of cart, we have:

a = 4 (aces)

t = 4 (three)

j = 4 (jacks)

And the total number of cards in the deck is

n = 52

So, the probability of drawing an ace as first cart is:

$p(a)=\frac{a}{n}=\frac{4}{52}=\frac{1}{13}=0.0769$

At the second drawing, the ace is not replaced within the deck. So the number of cards left in the deck is

$n-1=51$

Therefore, the probability of drawing a three at the 2nd draw is

$p(t)=\frac{t}{n-1}=\frac{4}{51}=0.0784$

Then, at the third draw, the previous 2 cards are not replaced, so there are now

$n-2=50$

cards in the deck. So, the probability of drawing a jack is

$p(j)=\frac{j}{n-2}=\frac{4}{50}=0.08$

Therefore, the total probability of drawing an ace, a three and then a jack is:

$p(atj)=p(a)\cdot p(j) \cdot p(t)=0.0769\cdot 0.0784 \cdot 0.08 =4.82\cdot 10^{-4}$

4 0

3 years ago

The graph of f(x) = 13* - 6 is shown below.g(x) is a transformation of f(x).

sergejj [24]

Answer:

your answer is proably 15 i dont really know

Step-by-step explanation:

8 0

3 years ago

What is the least 10-digit whole number?

Murljashka [212]

1000000000 because less than this number is 9 digits and more than this numbet is not least

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3 years ago