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3 years ago

11-15 pls help thanks

Mathematics

1 answer:

AlekseyPX3 years ago

7 0

What do I need to solve :)

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Let S be the universal set, where:

kolezko [41]

A' = {7,8,10}

B' = {1,3,59}

Given,

Let S be the universal set, where:

S = {1,2,3,4,5,6,7,8,9,10}

Let sets A and B be subsets of S, where:

Set A = {1,2,3,4,5,6,9}

Set B = {2,4,6,7,8,10}

We need to find the elements in the set (A') and (B)'

<h3>What is a complement of a set?</h3>

A compliment of a set D is denoted by D' or D bar.

It means the set contains all elements in the universal set U other than the elements in set D.

Example:

U = {1,2,3,4,5,6,7,8,9}

D = {1,2,3,4,5}

D' = {6,7,8,9}

Find A' and B'

S = {1,2,3,4,5,6,7,8,9,10}

A = {1,2,3,4,5,6,9}

B = {2,4,6,7,8,10}

A' = {7,8,10}

B' = {1,3,59}

Thus,

A' = {7,8,10}

B' = {1,3,59}

Learn more about the complement of sets here:

brainly.com/question/20319548

#SPJ1

3 0

2 years ago

2(x + 1 ) - 3(x + 5) ≥ 0

zaharov [31]

answer is x is less than or equal to -13

hope this helps:-)

3 0

3 years ago

Read 2 more answers

Jessica is going to a vegetable shop. She buys some lady fingers for $23, some potatoes for $30,

mario62 [17]

Answer:

$89

Step-by-step explanation:

add all the numbers together

6 0

3 years ago

Assume the hold time of callers to a cable company is normally distributed with a mean of 5.5 minutes and a standard deviation o

Ierofanga [76]

Answer:

The percent of callers are 37.21 who are on hold.

Step-by-step explanation:

Given:

A normally distributed data.

Mean of the data, $\mu$ = 5.5 mins

Standard deviation, $\sigma$ = 0.4 mins

We have to find the callers percentage who are on hold between 5.4 and 5.8 mins.

Lets find z-score on each raw score.

⇒ $z_1=\frac{x_1-\mu}{\sigma}$ ...raw score, $x_1$ = $5.4$

⇒ Plugging the values.

⇒ $z_1=\frac{5.4-5.5}{0.4}$

⇒ $z_1=-0.25$

For raw score 5.5 the z score is.

⇒ $z_2=\frac{5.8-5.5}{0.4}$

⇒ $z_2=0.75$

Now we have to look upon the values from Z score table and arrange them in probability terms then convert it into percentages.

We have to work with P(5.4<z<5.8).

⇒ $P(5.4$

⇒ $P(-0.25$

⇒ $P(z$

⇒ $z(1.5)=0.7734$ and $z(-0.25)=0.4013$ .<em>..from z -score table.</em>

⇒ $0.7734-0.4013$

⇒ $0.3721$

To find the percentage we have to multiply with 100.

⇒ $0.3721\times 100$

⇒ $37.21$ %

The percent of callers who are on hold between 5.4 minutes to 5.8 minutes is 37.21

4 0

3 years ago

there are 98 girls at Ross elementary there are 15 less boys then girls how many boys and girls are at Ross elementary

Luba_88 [7]

Answer: 98-15= 83 boys

8 0

3 years ago