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Komok [63]
3 years ago
12

during a soccer game Aaron scored 1 out of 2 goals attempted.Lauren scored 6 out of 8 goals attempted. write the fraction that r

epresents each and compare who was the most successful explain your answer
Mathematics
1 answer:
crimeas [40]3 years ago
5 0

Arron score 1/2

Lauren scored 6/8 or 3/4

3/4 is larger than 1/2 so Lauren scored on more of her attempted goals than Arron did which makes her more successful.

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Find the product of 2/3 of 30
weqwewe [10]
Hey there!

2/3 *30/1= 30/3= 10 and 3/3=1 so 2/1*10/1= 20
So the product is 20
6 0
4 years ago
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Simplify 3 sqrt 5x * 3 sqrt 25x^ 2 completely.​
tia_tia [17]

Answer:

=45x√5x

Step-by-step explanation:

The provided expression is: 3√5x×3√25x²

We can simplify 3√25x² because √25x²=5x

3×5x=15x

Combine with the rest of the expression through multiplication.

3√5x×15x

=45x√5x

The expression 3√5x×3√25x² is equivalent to 45x√5x.

The expected simplified answer is 45x√5x

8 1
3 years ago
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Help me figure out what X is
11Alexandr11 [23.1K]

8x+11=7x+26\\ 8x-7x=26-11\\ x=15

4 0
3 years ago
Activity
kobusy [5.1K]

a = (1+0.2)p

This equation is used to calculate the quantity of lemonade.

<u>Step-by-step explanation:</u>

Let the quality of lemonade in lans glass be 'a'

Quantity of lemonade in patricias glass = p ounces

Lan pours 20% more than Patricia

a = (1+0.2)p

This equation is used to calculate the quantity of lemonade

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3 years ago
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Forty percent of households say they would feel secure if they had $50,000 in savings. you randomly select 8 households and ask
Kay [80]

Answer:

Let X be the event of feeling secure after saving $50,000,

Given,

The probability of feeling secure after saving $50,000, p = 40 % = 0.4,

So, the probability of not  feeling secure after saving $50,000, q = 1 - p = 0.6,

Since, the binomial distribution formula,

P(x=r)=^nC_r p^r q^{n-r}

Where, ^nC_r=\frac{n!}{r!(n-r)!}

If 8 households choose randomly,

That is, n = 8

(a) the probability of the number that say they would feel secure is exactly 5

P(X=5)=^8C_5 (0.4)^5 (0.6)^{8-5}

=56(0.4)^5 (0.6)^3

=0.12386304

(b) the probability of the number that say they would feel secure is more than five

P(X>5) = P(X=6)+ P(X=7) + P(X=8)

=^8C_6 (0.4)^6 (0.6)^{8-6}+^8C_7 (0.4)^7 (0.6)^{8-7}+^8C_8 (0.4)^8 (0.6)^{8-8}

=28(0.4)^6 (0.6)^2 +8(0.4)^7(0.6)+(0.4)^8

=0.04980736

(c) the probability of the number that say they would feel secure is at most five

P(X\leq 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5)

=^8C_0 (0.4)^0(0.6)^{8-0}+^8C_1(0.4)^1(0.6)^{8-1}+^8C_2 (0.4)^2 (0.6)^{8-2}+8C_3 (0.4)^3 (0.6)^{8-3}+8C_4 (0.4)^4 (0.6)^{8-4}+8C_5(0.4)^5 (0.6)^{8-5}

=0.6^8+8(0.4)(0.6)^7+28(0.4)^2(0.6)^6+56(0.4)^3(0.6)^5+70(0.4)^4(0.6)^4+56(0.4)^5(0.6)^3

=0.95019264

8 0
3 years ago
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