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S_A_V [24]
3 years ago
15

Use the figure to find measures of the numbered angles.

Mathematics
2 answers:
cupoosta [38]3 years ago
7 0
1 and 4 are 113! And 2, and 3 are 67!
zysi [14]3 years ago
5 0
1 - 113 degrees

2 - 67 degrees

3 - 67 degrees

4 - 113 degrees












Hope this helped :)
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Answer to get 20 points and brainleiest
ddd [48]

Answer:

x=10/7

Step-by-step explanation:

8 0
2 years ago
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Write a problem that includes multiplying decimals . Explain how you know where to place the decimal in the product.
loris [4]
Maya is making dream catchers with her family. Each dream catcher takes 1.53 feet of string. She has 6 people in her family. How much string does Maya need total?

1.53*6=9.18

You know where to put the decimal because you count how many places in the decimal is for the two numbers being multiplied (the only decimal in the two numbers being multiplied is in the number 1.53 and that’s is two places in) and then go that many places in on the product (918 two places in from then right would be 9.18)
4 0
3 years ago
Find the probability of getting four consecutive aces when four cards are drawn without replacement from a standard deck of 52 p
posledela

Answer:

<em>P=0.0000037</em>

<em>P=0.00037%</em>

Step-by-step explanation:

<u>Probability</u>

A standard deck of 52 playing cards has 4 aces.

The probability of getting one of those aces is

\displaystyle \frac{4}{52}=\frac{1}{13}

Now we got an ace, there are 3 more aces out of 51 cards.

The probability of getting one of those aces is

\displaystyle \frac{3}{51}=\frac{1}{17}

Now we have 2 aces out of 50 cards.

The probability of getting one of those aces is

\displaystyle \frac{2}{50}=\frac{1}{25}

Finally, the probability of getting the remaining ace out of the 49 cards is:

\displaystyle \frac{1}{49}

The probability of getting the four consecutive aces is the product of the above-calculated probabilities:

\displaystyle P= \frac{1}{13}\cdot\frac{1}{17}\cdot\frac{1}{27}\cdot\frac{1}{49}

\displaystyle P= \frac{1}{270,725}

P=0.0000037

P=0.00037%

3 0
2 years ago
Suppose that each child born is equally likely to be a boy or a girl. Consider a family with exactly three children. Let BBG ind
Gemiola [76]

Answer:

(a)

S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}

(b)

i.

1\ girl = \{GBB, BBG, BGB\}

P(1\ girl) = 0.375

ii.

Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}

P(Atleast\ 2 \ girls) = 0.5

iii.

No\ girl = \{BBB\}

P(No\ girl) = 0.125

Step-by-step explanation:

Given

Children = 3

B = Boys

G = Girls

Solving (a): List all possible elements using set-roster notation.

The possible elements are:

S = \{GGG, GGB, GBG, GBB, BBG, BGB, BGG, BBB\}

And the number of elements are:

n(S) = 8

Solving (bi) Exactly 1 girl

From the list of possible elements, we have:

1\ girl = \{GBB, BBG, BGB\}

And the number of the list is;

n(1\ girl) = 3

The probability is calculated as;

P(1\ girl) = \frac{n(1\ girl)}{n(S)}

P(1\ girl) = \frac{3}{8}

P(1\ girl) = 0.375

Solving (bi) At least 2 are girls

From the list of possible elements, we have:

Atleast\ 2 \ girls = \{GGG, GGB, GBG, BGG\}

And the number of the list is;

n(Atleast\ 2 \ girls) = 4

The probability is calculated as;

P(Atleast\ 2 \ girls) = \frac{n(Atleast\ 2 \ girls)}{n(S)}

P(Atleast\ 2 \ girls) = \frac{4}{8}

P(Atleast\ 2 \ girls) = 0.5

Solving (biii) No girl

From the list of possible elements, we have:

No\ girl = \{BBB\}

And the number of the list is;

n(No\ girl) = 1

The probability is calculated as;

P(No\ girl) = \frac{n(No\ girl)}{n(S)}

P(No\ girl) = \frac{1}{8}

P(No\ girl) = 0.125

7 0
3 years ago
Which statement is an example of the Addition Property of Equality?
weqwewe [10]
The answer would be p + s = q + s      ...so it would be B.
6 0
2 years ago
Read 2 more answers
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