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2 years ago

I'm using equations what is the solution of this email equation 6d=108 what do d=

Mathematics

1 answer:

bonufazy [111]2 years ago

3 0

D = 18
you divide 108 by 6
and that's how you get 18 as an answer

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Variable y varies directly with variable x, and y = 48 when x = 8.

scZoUnD [109]

Y/x=k 48/8=6=k k is the constant y/4=6 y=24

3 0

2 years ago

What values are solutions of 0 =2xexponent of 2 -6x-8

Svetlanka [38]

Answer:

x = -1 and x = 4

Step-by-step explanation:

Firstly solve the quadratic expression of

2x^2-6x-8 = 0....it gives you the two brackets of.....(2x-8)(x+1)=0

then...you equate the two brackets to 0.separately

2x-8 = 0....which gives you...x = 4...as the first solution..then..;

x+1 = 0....which gives you...x = -1...as the second solution

3 0

2 years ago

Evaluate each expression. 3^-2

Rus_ich [418]

9 is the answer negatives cancel out creating positve

7 0

2 years ago

Read 2 more answers

Bridget rolls a fair dice 84 times.<br> How many times would Bridget expect to roll an odd number?

cluponka [151]

Answer:

42 times

Step-by-step explanation:

Since there are 6 sides on a fair dice, each having a number 1 thru 6 with no numbers repeating and half those numbers being odd, Bridget can expect to roll an odd number 84*(1/2) times. So our answer is 42.

Also, if you could please mark this "brainliest" (the crown) it would help a lot!

6 0

2 years ago

Situation D: Suppose that, in a one-minute period during an electrical storm, the number of lightning strikes on a radar antenna

Crank

Answer:

21.77% probability that the antenna will be struck exactly once during this time period.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

In this question:

$\mu = 2.40$

Find the probability that the antenna will be struck exactly once during this time period.

This is P(X = 1).

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-2.40}*2.40^{1}}{(1)!} = 0.2177$

21.77% probability that the antenna will be struck exactly once during this time period.

3 0

2 years ago