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3 years ago

9

4x-6 + 2x = 18 What’s the answer

1 answer:

solmaris [256]3 years ago

6 0

Answer:

x=4

Step-by-step explanation:

4x-6 + 2x = 18

Combine like terms

6x -6 = 18

Add 6 to each side

6x-6+6 =18+6

6x = 24

Divide each side by 6

6x/6 = 24/6

x =4

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Weary of the low turnout in student elections, a college administration decides to choose an SRS of three students to form an ad

-Dominant- [34]

Answer:

$P(ABC) = 0.110592$

$P(ABC^c) = 0.119808$

$P(AB^cC) = 0.119808$

$P(A^cBC) = 0.119808$

$P(AB^cC^c) = 0.129792$

$P(A^cBC^c) = 0.129792$

$P(A^cB^cC) = 0.129792$

$P(A^cB^cC^c) = 0.140608$

Step-by-step explanation:

Given

$P(A) = P(B) = P(C) = 48\%$

Convert the probability to decimal

$P(A) = P(B) = P(C) = 0.48$

Solving (a): P(ABC)

This is calculated as:

$P(ABC) = P(A) * P(B) * P(C)$

This gives:

$P(ABC) = 0.48*0.48*0.48$

$P(ABC) = 0.110592$

Solving (b): $P(ABC^c)$

This is calculated as:

$P(ABC^c) = P(A) * P(B) * P(C^c)$

In probability:

$P(C^c) = 1 - P(C)$

So, we have:

$P(ABC^c) = P(A) * P(B) * (1 - P(C))$

$P(ABC^c) = 0.48 * 0.48 * (1 - 0.48)$

$P(ABC^c) = 0.48 * 0.48 * 0.52$

$P(ABC^c) = 0.119808$

Solving (c): $P(AB^cC)$

This is calculated as:

$P(AB^cC) = P(A) * P(B^c) * P(C)$

$P(AB^cC) = P(A) * [1 - P(B)] * P(C)$

$P(AB^cC) = 0.48 * (1 - 0.48)* 0.48$

$P(AB^cC) = 0.48 * 0.52* 0.48$

$P(AB^cC) = 0.119808$

Solving (d): $P(A^cBC)$

This is calculated as:

$P(A^cBC) = P(A^c) * P(B) * P(C)$

$P(A^cBC) = [1-P(A)] *P(B) * P(C)$

$P(A^cBC) = (1 - 0.48)* 0.48 * 0.48$

$P(A^cBC) = 0.52* 0.48 * 0.48$

$P(A^cBC) = 0.119808$

Solving (e): $P(AB^cC^c)$

This is calculated as:

$P(AB^cC^c) = P(A) * P(B^c) * P(C^c)$

$P(AB^cC^c) = P(A) * [1-P(B)] * [1-P(C)]$

$P(AB^cC^c) = 0.48 * [1-0.48] * [1-0.48]$

$P(AB^cC^c) = 0.48 * 0.52*0.52$

$P(AB^cC^c) = 0.129792$

Solving (f): $P(A^cBC^c)$

This is calculated as:

$P(A^cBC^c) = P(A^c) * P(B) * P(C^c)$

$P(A^cBC^c) = [1-P(A)] * P(B) * [1-P(C)]$

$P(A^cBC^c) = [1-0.48] * 0.48 * [1-0.48]$

$P(A^cBC^c) = 0.52 * 0.48 * 0.52$

$P(A^cBC^c) = 0.129792$

Solving (g): $P(A^cB^cC)$

This is calculated as:

$P(A^cB^cC) = P(A^c) * P(B^c) * P(C)$

$P(A^cB^cC) = [1-P(A)] * [1-P(B)] * P(C)$

$P(A^cB^cC) = [1-0.48] * [1-0.48] * 0.48$

$P(A^cB^cC) = 0.52 * 0.52 * 0.48$

$P(A^cB^cC) = 0.129792$

Solving (h): $P(A^cB^cC^c)$

This is calculated as:

$P(A^cB^cC^c) = P(A^c) * P(B^c) * P(C^c)$

$P(A^cB^cC^c) = [1-P(A)] * [1-P(B)] * [1-P(C)]$

$P(A^cB^cC^c) = [1-0.48] * [1-0.48] * [1-0.48]$

$P(A^cB^cC^c) = 0.52*0.52*0.52$

$P(A^cB^cC^c) = 0.140608$

5 0

2 years ago

Stanley Bank offers a money market account that earns 1.85% compounded continuously: a) If $10,000 is invested in this type of a

Alexxx [7]

Answer:

a) $1056.33 b) 23 years

Step-by-step explanation:

a) 10000(1+1.85/100)^3=10565.33 (2d.p.)

b) let x be the no. of years

15000 = 10000(1+1.85/100)^x

1.5 = 1.0185^x

ln both sides

ln 1.5 = x ln 1.0185

x = ln 1.5/ln 1.0185

=22.11

=23 years (rounded up)

5 0

2 years ago

Expressions equivalent to -4(x-9)

lisov135 [29]

-4x +36 I am pretty sure

3 0

2 years ago

A recipe calls for 2Start Fraction 3 over 4 End Fraction cups of sugar. How many fourths is that? A. 3 B. 8 C. 11 D. 14

pickupchik [31]

I’m pretty sure that would be B

4 0

2 years ago

Read 2 more answers

Newborn males have weights with a mean of 3272.8 g and a standard deviation of 660.2 g. Newborn females have weights with a mean

Maksim231197 [3]

The correct option is: a female who weighs 1500 g

<em><u>Explanation</u></em>

<u>Formula for finding the z-score</u> is: $z= \frac{X-\mu}{\sigma}$

Newborn males have weights with a mean $(\mu)$ of 3272.8 g and a standard deviation $(\sigma)$ of 660.2 g.

So, the z-score for the newborn male who weighs 1500 g will be.......

$z(X=1500)=\frac{1500-3272.8}{660.2}=-2.685... \approx -2.69$

According to the normal distribution table, $P(z=-2.69)=0.0036 = 0.36\%$

Now, newborn females have weights with a mean $(\mu)$ of 3037.1 g and a standard deviation $(\sigma)$ of 706.3 g.

So, the z-score for the newborn female who weighs 1500 g will be.......

$z(X=1500)=\frac{1500-3037.1}{706.3}=-2.176... \approx -2.18$

According to the normal distribution table, $P(z=-2.18)=0.0146 = 1.46\%$

As we can see that the <u>probability that a newborn female has weight of 1500 g is greater than newborn male</u>, so a newborn female has the weight of 1500 g that is more extreme relative to the group from which he came.

5 0

2 years ago