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harina [27]
3 years ago
15

How many Line segment in square, rectangle and circles

Mathematics
2 answers:
Lena [83]3 years ago
8 0
Square-4
Rectangle-4
Circle-1
nlexa [21]3 years ago
5 0

Answer:

You need four of them in order to construct a square. and rectangle.A circle doesn't have line segments. It is a curve, not made up of straight lines.

Step-by-step explanation:

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Function 1 is defined by the equation p= --3/2r - 5. function 2 is defined by the following table. Which function has a greater
Rus_ich [418]
0-5. Here’s your answer
6 0
2 years ago
HELP NEED ASAP FOR TOMORROW ITS FOR MY HOMEWORK!! ILL MARK BRAINLIEST IF RIGHT!!​
Slav-nsk [51]

Answer:

  y ≥ -x +2

Step-by-step explanation:

The solid line has a slope of -1 and a y-intercept of 2, so its equation in slope-intercept form is ...

  y = -x +2

The shaded area is above this line, and the line is part of the solution set, so we want an inequality that has "y" and the comparison symbol in this order: "y ≥" or "≤ y".

We already have an equation with "y" on the left, above, so we just need to introduce the comparison symbol:

  y ≥ -x +2

Another way to write this is ...

  x + y ≥ 2

4 0
3 years ago
What’s the probability of getting each card out of a deck?
VMariaS [17]

Answer:

a. 1/13

b. 1/52

c. 2/13

d. 1/2

e. 15/26

f. 17/52

g. 1/2

Step-by-step explanation:

a. In a deck of cards, there are 4 suits and each of them has a 7. Therefore, the probability of drawing a 7 is:

P(7) = 4/52 = 1/13

b. There is only one 6 of clubs, therefore, the probability of drawing a 6 of clubs is:

P(6 of clubs) = 1/52

c. There 4 fives (one for each suit) and 4 queens in a deck of cards. Therefore, the probability of drawing a five or a queen​​​​​​​​​​​ is:

P(5 or Q) = P(5) + P(Q)

= 4/52 + 4/52

= 1/13 + 1/13

P(5 or Q) = 2/13

d. There are 2 suits that are black. Each suit has 13 cards. Therefore, there are 26 black cards. The probability of drawing a black card is:

P(B) = 26/52 = 1/2

e. There are 2 suits that are red. Each suit has 13 cards. Therefore, there are 26 red cards. There are 4 jacks. Therefore:

P(R or J) = P(R) + P(J)

= 26/52 + 4/52

= 30/52

P(R or J) = 15/26

f. There are 13 cards in clubs suit and there are 4 aces, therefore:

P(C or A) = P(C) + P(A)

= 13/52 + 4/52

P(C or A) = 17/52

g. There are 13 cards in the diamonds suit and there are 13 in the spades suit, therefore:

P(D or S) = P(D) + P(S)

= 13/52 + 13/52

= 26/52

P(D or S) = 1/2

6 0
3 years ago
Anderson won $10000 in a Lotto competition.
kozerog [31]

Answer:

jimmy got 30,000

Step-by-step explanation:

10,000

the wife got 60% = 60,000

jimmy got 3x prince = 30,000

prince = 10,000

8 0
3 years ago
The table gives the relative frequencies of recipes that contains sugar and salt, contain at least one of those ingredients, or
tigry1 [53]

<u>Answer-</u>

<em>The probability that a randomly selected recipe does not contain sugar, given that it contains salt is 22.4%</em>

<u>Solution-</u>

The given table in the link shows the relative frequencies of recipes that contains sugar and salt, or contains at least one of those ingredients, or contains neither of those ingredients.

We have to find the conditional probability that the recipe doesn't contain sugar, given that it contains salt.

We know that, the conditional probability of occurrence of A given that B occurs is,

P(A|B)=\frac{P(A\ and\ B)}{P(B)} =\frac{P(A\bigcap B)}{P(B)}

P(\text{Doesn't contain sugar}\ |\ \text{Contains salt})=\frac{P(\text{Doesn't contain sugar}\ and\ \text{Contains salt})}{P(\text{Contains salt})}


P(\text{Doesn't contain sugar}\ and\ \text{Contains salt})=\frac{0.15}{1} =0.15\\P(\text{Contains salt})=\frac{0.67}{1}=0.67

Putting these values,

P(\text{Doesn't contain sugar}\ |\ \text{Contains salt})=\frac{0.15}{0.67} =0.224=22.4\%

5 0
3 years ago
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