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3 years ago

My question is where is the nearest fraction store i need to know

Mathematics

1 answer:

Alla [95]3 years ago

5 0

Erm.... Just down the high street, two thirds of a mile past the parking lot. Its door number is 1 and a quarter. XD

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Olivia has a points card for a movie theater. She receives 25 rewards points just for signing up. She earns 7.5 points for each

miv72 [106K]

Answer:

0 visits

Step-by-step explanation:

According to the scenario, calculation of the given data are as follows,

Rewards for signing up = 80 points

Reward per visit = 2.5 points

Total points needed = 105

Let, number of visits = x

So, we can calculate number of visits by using following formula,

80 + (2.5 × X) = 105

2.5X = 105 - 80

2.5X = 25

X = 25 ÷ 2.5

X = 10 visits.

Hence, Olivia needs to make 10 visits in order to earn a free movie ticket

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

HOW MANY TIMES DOES 6 GO INTO 25

Sveta_85 [38]

Answer:

25/6 = 4 with a remainder of 1

8 0

3 years ago

Read 2 more answers

Consider two independent tosses of a fair coin. Let A be the event that the first toss results in heads, let B be the event that

aliina [53]

Answer with Step-by-step explanation:

We are given that two independent tosses of a fair coin.

Sample space={HH,HT,TH,TT}

We have to find that A, B and C are pairwise independent.

According to question

A={HH,HT}

B={HH,TH}

C={TT,HH}

$A\cap B=$ {HH}

$B\cap C$ ={HH}

$A\cap C$ ={HH}

P(E)= $\frac{number\;of\;favorable\;cases}{total\;number\;of\;cases}$

Using the formula

Then, we get

Total number of cases=4

Number of favorable cases to event A=2

$P(A)=\frac{2}{4}=\frac{1}{2}$

Number of favorable cases to event B=2

Number of favorable cases to event C=2

$P(B)=\frac{2}{4}=\frac{1}{2}$

$P(C)=\frac{2}{4}=\frac{1}{2}$

If the two events A and B are independent then

$P(A)\cdot P(B)=P(A\cap B)$

$P(A\cap)=\frac{1}{4}$

$P(B\cap C)=\frac{1}{4}$

$P(A\cap C)=\frac{1}{4}$

$P(A)\cdot P(B)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$

$P(B)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(C)=\frac{1}{4}$

$P(A)\cdot P(B)=P(A\cap B)$

Therefore, A and B are independent

$P(B)\cdot P(C)=P(B\cap C)$

Therefore, B and C are independent

$P(A\cap C)=P(A)\cdot P(C)$

Therefore, A and C are independent.

Hence, A, B and C are pairwise independent.

6 0

3 years ago

Ryan is trying a low-carbohydrate diet. He would like to keep the amount of carbs consumed in grams between the levels shown in

pentagon [3]

Compound inequality combines multiple single inequalities.

The solution and interpretation of the inequality is:

$x > 50$ and $x < 150$ . Ryan needs to consume more than 50 grams of carbohydrates, but less than 150 grams of carbohydrates.

We have:

$110 < 2x + 10$ and $2x + 10< 310$

Solve for x in both inequalities

$110 < 2x + 10$

$110 -10 < 2x$

$100 < 2x$

Divide both sides by 2

$50 < x$

Rewrite as:

$x > 50$

Also, we have:

$2x + 10< 310$

$2x< 310 - 10$

$2x< 300$

Divide by 2

$x < 150$

So, the solutions to the inequalities are:

$x > 50$ and $x < 150$

This means that Ryan needs to consume more than 50 but less than 150 grams of carbohydrates.

Hence, (a) is correct

Read more about compound inequalities at:

brainly.com/question/13290962

8 0

3 years ago

I will mark you as brainlis if you andwer this all.

dimaraw [331]

Answer: See below

Step-by-step explanation:

27. -(a-3)

28. (b-1)(b+3)

29. (c+4)(c+5)

30. d(d+5)

31. -(3/4)(2e-5)

Sorry - I don't have time to enter the details. Look for areas where the expressions can be factored in a manner that forms as many equivalent expressions in both the numerator and denominator.

For example: In problem 30:

(5d-20)/(d^2+d-20) * [??]/20d = 1/4

Factor:

(5(d-4)) d(d+5) = 1/4

(d-4)(d+5) 20d

The (d-4), d+5, and d terms cancel, leaving

5/20 = 1/4

7 0

2 years ago