Please help!!10 points!!

Mathematics

1 answer:

ANTONII [103]2 years ago

3 0

Answer:

the hell Well fist off you need to be able to graph to if you have a phone and look it up.

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-3(4x+3)+4(6x+1)43 please solve the problem fully

e-lub [12.9K]

Answer and Step-by-step explanation:

First, distribute the numbers outside of the parenthesis to the numbers on the inside.

-12x - 9 + 24x + 4 = 43

Combine like terms.

12x - 5 = 43

Add 5 to both sides.

12x = 48

Divide by 12 on both sides of the equation.

x = 3 <- This is the answer.

#teamtrees #PAW (Plant And Water)

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3 years ago

A New cell phone cost $225. The accessories cost 13% of the cell phone price. What is the cost of the accessories?

Anvisha [2.4K]

$29.25 use a calculator, 225 - 13% = 195.75, 225 - 195.75 = 29.25

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3 years ago

mercury orbits the sun in 87 and 24/50 venus orbits the sun in 224 and 7/10 earthdays and mars orbits the sun in 686 and 49/50 d

slega [8]

Mercury: 87 24/25
Venus 224 7/10
Mars 686 49/50

All we need to do is to convert fractions so they have a denominator of 10, 100, 1000 etc.

87 24/25 = 87 96/100
224 7/10 stays the same
686 49/50 = 686 98/100

Now we can easily convert them into decimals:

Mercury: 87.96
Venus: 224.7
Mars: 686.98

7 0

3 years ago

assume that women's heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. find the

drek231 [11]

Answer:

65.3658 inches

Step-by-step explanation:

Let X be the height of a woman randomly choosen. We know tha X have a mean of 63.6 inches and a standard deviation of 2.5 inches. For an x value, the related z-score is given by z = (x-63.6)/2.5. We are looking for a value $x_{0}$ such that $P(X < x_{0}) = 0.76$ , but, $0.76 = P(X < x_{0}) = P((X-63.6)/2.5 < (x_{0}-63.6)/2.5) = P(Z < (x_{0}-63.6)/2.5)$ , i.e., $(x_{0}-63.6)/2.5$ is the 76th percentile of the standard normal distribution. So, $(x_{0}-63.6)/2.5 = 0.7063$ , $x_{0} =63.6+(2.5)(0.7063) = 65.3658$ . Therefore, the height of a woman who is at the 76th percentile is 65.3658 inches.