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3 years ago

Find the measure of the given arc in Circle C.

Mathematics

1 answer:

IRINA_888 [86]3 years ago

4 0

Answer:

AE=90°

AED=140°

DBA=220°

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What is the Range of this data

musickatia [10]

The range of this data is 65 since the minimum is 20 and the maximum is 85(if its counting by 5s) then you just subtract and you have the range.

8 0

3 years ago

I need help with number 5

Tju [1.3M]

Answer:

B. BD = 9

Step-by-step explanation:

Simply set the 2 equations equal to each other (both are congruent due to perpendicular bisector):

6x + 3 = 3x = 6

3x + 3 = 6

3x = 3

x = 1

Then substitute 1 for x in BD

6(1) + 3

6 + 3

BD = 9

7 0

3 years ago

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What is the fraction of 2.6

Solnce55 [7]

2 3/5 or 2 6/10 is the answer

6 0

3 years ago

Read 2 more answers

a vido game system is being sold at a 4% discount. if the discount results in a saving of$11,96, what is the original price of t

kramer

Answer:

$299

Step-by-step explanation:

Divide the discount amount through the discount percentage and multiply with a hundred percent, resulting in the full price.

($11,96)/4*100=$299

8 0

3 years ago

In a certain Algebra 2 class of 30 students, 19 of them play basketball and 12 of them play baseball. There are 8 students who p

Alenkinab [10]

Answer:

Probability that a student chosen randomly from the class plays basketball or baseball is $\frac{23}{30}$ or 0.76

Step-by-step explanation:

Given:

Total number of students in the class = 30

Number of students who plays basket ball = 19

Number of students who plays base ball = 12

Number of students who plays base both the games = 8

To find:

Probability that a student chosen randomly from the class plays basketball or baseball=?

Solution:

$P(A \cup B)=P(A)+P(B)-P(A \cap B)$ ---------------(1)

where

P(A) = Probability of choosing a student playing basket ball

P(B) = Probability of choosing a student playing base ball

P(A \cap B) = Probability of choosing a student playing both the games

Finding P(A)

P(A) = $\frac{\text { Number of students playing basket ball }}{\text{Total number of students}}$

P(A) = $\frac{19}{30}$ --------------------------(2)

Finding P(B)

P(B) = $\frac{\text { Number of students playing baseball }}{\text{Total number of students}}$

P(B) = $\frac{12}{30}$ ---------------------------(3)

Finding $P(A \cap B)$

P(A) = $\frac{\text { Number of students playing both games }}{\text{Total number of students}}$

P(A) = $\frac{8}{30}$ -----------------------------(4)

Now substituting (2), (3) , (4) in (1), we get

$P(A \cup B)= \frac{19}{30} + \frac{12}{30} -\frac{8}{30}$

$P(A \cup B)= \frac{31}{30} -\frac{8}{30}$

$P(A \cup B)= \frac{23}{30}$

7 0

3 years ago