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Aliun [14]
4 years ago
5

Help please 2+2-1-1+6=x

Mathematics
2 answers:
natita [175]4 years ago
8 0

Answer:

2+2-1-1+6=x : x = 8

Step-by-step explanation:

2+2-1-1+6=x\\

Switch sides:

x=2+2-1-1+6

Add/Subtract the numbers: 2+2-1-1+6=8

x = 8

Hope I helped, if so may I get brainliest and a thanks?

Thank you, have a good day! =)

Anestetic [448]4 years ago
6 0

Answer:

8

Step-by-step explanation:

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18-2x+10=12x solve for x
Scorpion4ik [409]

Answer:

x=2

Step-by-step explanation:

18-2x+10=12x

-2x+18+10=12x

-2x+28=12x

subtract 28 from both sides= -2x=12x-28

subtract 12x from both sides= -14x=-28

divide both sides by -14= x=2

x=2 is your final answer

8 0
3 years ago
Read 2 more answers
105. Suppose that the probability that an adult in America will watch the Super Bowl is 40%. Each person is considered independe
Ainat [17]

Answer:

a. X is the number of adults in America that need to be surveyed until finding the first one that will watch the Super Bowl.

b. X can take any integer that is greater than or equal to 1. \rm X\in \mathbb{Z}^{+}.

c. \rm X \sim NB(1, 0.40).

d. E(\rm X) = 2.5.

e. P(\rm X = 7) = 0.0187.

f. P(\text{X} = 3) +P(\text{X} = 4) = 0.230.

Step-by-step explanation:

<h3>a.</h3>

In this setting, finding an adult in America that will watch the Super Bowl is a success. The question assumes that the chance of success is constant for each trial. The question is interested in the number of trials before the first success. Let X be the number of adults in America that needs to be surveyed until finding the first one who will watch the Super Bowl.

<h3>b.</h3>

It takes at least one trial to find the first success. However, there's rare opportunity that it might take infinitely many trials. Thus, X may take any integer value that is greater than or equal to one. In other words, X can be any positive integer: \rm X\in \mathbb{Z}^{+}.

<h3>c.</h3>

There are two discrete distributions that may model X:

  • The geometric distribution. A geometric random variable measures the number of trials before the first success. This distribution takes only one parameter: the chance of success on each trial.
  • The negative binomial distribution. A negative binomial random variable measures the number of trials before the r-th success. This distribution takes two parameters: the number of successes r and the chance of success on each trial p.

\rm NB(1, p) (note that r=1) is equivalent to \sim Geo(p). However, in this question the distribution of \rm X takes two parameters, which implies that \rm X shall follow the negative binomial distribution rather than the geometric distribution. The probability of success on each trial is 40\% = 0.40.

\rm X\sim NB(1, 0.40).

<h3>d.</h3>

The expected value of a negative binomial random variable is equal to the number of required successes over the chance of success on each trial. In other words,

\displaystyle E(\text{X}) = \frac{r}{p} = \frac{1}{0.40} = 2.5.

<h3>e.</h3>

P(\rm X = 7) = 0.0187.

Some calculators do not come with support for the negative binomial distribution. There's a walkaround for that as long as the calculator supports the binomial distribution. The r-th success occurs on the n-th trial translates to (r-1) successes on the first (n-1) trials, plus another success on the n-th trial. Find the chance of (r-1) successes in the first (n-1) trials and multiply that with the chance of success on the n-th trial.

<h3>f.</h3>

P(\text{X} = 3)+P(\text{X} = 4) = 0.230.

3 0
3 years ago
What is the value of this number in decimal form?<br> Five and three hundred eight thousandths.
Rufina [12.5K]
5.308 would be the correct answer
4 0
3 years ago
Select the expressions that are equivalent to 8(6m).
Bas_tet [7]

\textsf{Hey there!}

\text{Select the expressions that are equicalent to 8(6m)}\downarrow

\text{We DON'T have any like terms so we CAN'T pair them off!}

\mathsf{8(6m) \rightarrow 8 \times 6 \times \ m}\\\mathsf{8\times 6 = 48}}\\\mathsf{48\times\  m = 48m}\\\\\\\boxed{\boxed{\mathsf{Answer: \bf{48m}}}}\checkmark

\textsf{Good luck on your assignment and enjoy your day!}

~\dfrac{\frak{LoveYourselfFirst}}{:)}

4 0
3 years ago
while visiting a friend's home, i saw kittens and children playing in the backyard. counting heads, i got 11. counting feet, i g
trapecia [35]

Answer:

There are 5 cats and 6 children

Step-by-step explanation:

Let x = number of cats

Let y= number of children

Counting heads gives 11

X+Y = 11 .....eq1

Counting feet gives 32

Cats has 4legs,child has 2 legs

4X + 2Y=32 ....eq2

X = 11 - Y

Put in eq2

4(11-Y) + 2Y = 32

44 - 4Y + 2Y=32

12=2Y

Y=12/2=6

Y=6= Children

Put Y=6 in eq2

X + 6=11

X=11-6=5

X= 5 = cats

7 0
3 years ago
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