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Mashutka [201]
3 years ago
8

Two numbers have a sum of 12 and a difference of 4. What is the smaller of the two numbers?

Mathematics
1 answer:
stiv31 [10]3 years ago
5 0
Answer: 8 and 4 hehe
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Write a paragraph comparing the two classes' semester grades. Be sure to compare the extremes, the quartiles, the medians, and t
Gnom [1K]

Answer:

Second class have higher marks and greater spread.

Step-by-step explanation:

First box plot  represents class first. From the first box plot, we get

\text{Minimum value }= 53,Q_1=62,Median=80,Q_3=86,\text{Maximum value }= 89

Range=Maximum-Minimum=89-53=36

IQR=Q_3-Q-1=86-62=24

Second box plot  represents class second. From the second box plot, we get

\text{Minimum value }= 56,Q_1=62,Median=74,Q_3=89,\text{Maximum value }= 96

Range=Maximum-Minimum=96-56=40

IQR=Q_3-Q-1=89-62=27

First class has greater minimum value, first quartile of both classes are same, second class has greater median, first class has greater third quartile and first class has greater maximum value. It means second class have higher marks but class first have less variation.

Second class has greater range and greater inter quartile range. It means data of second class has greater spread.

Therefore, second class have higher marks and greater spread.

3 0
3 years ago
A bicycle lock has a four-digit code. The possible digits, 0 through 9, cannot be repeated. What is the probability that the loc
kozerog [31]
First let's find the number of the elements in the sample space, that is the total number of codes that can be produced.

The first digit is any of {0, 1, 2...,9}, that is 10 possibilities
the second digit is any of the remaining 9, after having picked one. 
and so on...

so in total there are 10*9*8*7 = 5040 codes.

a. What is the probability that the lock code will begin with 5?

Lets fix the first number as 5. Then there are 9 possibilities for the second digit, 8 for the third on and 7 for the last digit.

Thus, there are 1*9*8*7=504 codes which start with 5.

so 

P(first digit is five)=\frac{n(first -digit- is- 5)}{n(all-codes)}= \frac{1*9*8*7 }{10*9*8*7 }= \frac{1}{10}=0.1

b. <span>What is the probability that the lock code will not contain the number 0? 

from the set {0, 1, 2...., } we exclude 0, and we are left with {1, 2, ...9}

from which we can form in total 9*8*7*6 codes which do not contain 0.

P(codes without 0)=n(codes without 0)/n(all codes)=(9*8*7*6)/(10*9*8*7)=6/10=0.6

Answer:

0.1 ; 0.6</span>
4 0
3 years ago
Read 2 more answers
the graph of f(x) is shown below. what are the values of a, b, c, and d in the definition of f(x) below? (graph and definition o
Otrada [13]

Answer:

Step-by-step explanation:

For the given piecewise function,

Let the equation of the segment AB is,

y = ax + 7 where x < -2

Since a point (-2, 3) lies on the given line,

3 = a(-2) + 7

3 - 7 = -2a

a = 2

Therefore, f(x) = 2x + 7 when x < -2

For the segment BC,

f(x) = 3 where -2 ≤ x ≤ 2

For the segment CD,

f(x) = bx + 9

Since a point (2, 3) lies on this segment,

f(2) = 2b + 9 = 3

2b = 3 - 9

b = -3

Therefore, equation of segment CD will be,

f(x) = (-3)x + 9   where x > 2

7 0
3 years ago
Please help if u know the answers quarter ends in two days !!!
MatroZZZ [7]

Answer:

5A) radius = 9 ft

5B) diameter = 18 ft

5C) fencing length = 56.52 ft

5D) 6.35 days

Step-by-step explanation:

Radius is given when it is 9 ft of rope (outside to center of circle)

Diameter is two times radius = 18.

Circumference(fence length) = 2 * 3.14 * 9 = 56.52.

Area circle = 3.14*(9*9) = 254.34. 254.34/40 = 6.35

If my answer is incorrect, pls correct me!

If you like my answer and explanation, mark me as brainliest!

-Chetan K

7 0
3 years ago
Suppose a compact disk​ (CD) you just purchased has 1515 tracks. After listening to the​ CD, you decide that you like 66 of the
stepladder [879]

Answer:

(A) 0.4196

(B) 0.2398

(C) 0.0020

Step-by-step explanation:

Given,

Total songs = 15,

Liked songs = 6,

So, not liked songs = 15 - 6 = 9

If any 5 songs are played,

Then the total number of ways =  ^{15}C_5

(A) Number of ways of choosing 2 liked songs = ^6C_2\times ^9C_3

Since,

\text{Probability}=\frac{\text{Favourable outcomes}}{\text{Total outcomes}}

Thus, the probability of choosing 3 females and 2 males = \frac{ ^6C_2\times ^9C_3}{^{15}C_5}

=\frac{\frac{6!}{2!4!}\times \frac{9!}{3!6!}}{\frac{15!}{10!5!}}

= 0.4196

Similarly,

(B)

The probability of choosing 3 liked songs = \frac{ ^6C_3\times ^9C_2}{^{15}C_5}

=\frac{\frac{6!}{3!3!}\times \frac{9!}{2!7!}}{\frac{15!}{10!5!}}

= 0.2398

(C)

The probability of choosing 5 liked songs = \frac{ ^6C_5\times ^9C_0}{^{15}C_5}

=\frac{\frac{6!}{5!1!}}{\frac{15!}{5!10!}}

≈ 0.0020

3 0
3 years ago
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