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morpeh [17]
3 years ago
15

Can anyone help me with the 7th and 8th question which I have taken a bad photo of?

Mathematics
1 answer:
Vsevolod [243]3 years ago
7 0
<u>Question 7:

Define x :
</u>
Let the digit in the ones place be x.
The digit in the tens place is  x+2.

The sum of the digit and its reverse is 66

<u>Find x:</u>

10(x + 2) + x + 10x + (x + 2) = 66
10x  + 20 + x + 10x + x + 2 = 66
22x + 22 = 66
22x = 44
x = 2

<u>Find the number:</u>

The number in the ones place = x= 2
The number in the tens place = x = 2 + 2 = 4

Answer : The number is 42.

<u>Question 8:</u>

<u>Define x:</u>

Let the digit in the tens place be x.
The digit in the ones place is x+7.

The number  = 10x + x + 7 = 11x + 7

If the digit interchange,
the number = 10(x + 7) + x = 10x + 70 + x = 11x + 70

<u>Find x:
</u>
9(11x + 7) = 2(11x + 70)
99x + 63 = 22x + 140
99x - 22x = 140 - 63
77x = 77
x = 1

<u>Find the number:
</u>
x = 1
x + 7 = 1 + 7 = 8
So the number is 18

Answer: The number is 18.

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there are 135 teams in a soccer tournament. each team has 32 players. how many soccer players in all are in the league
ozzi
Answer: 4,320

Explanation: Since there are 35 players on a team, you must find out how many players there would be if it was 35 times more so it would be 135 x 32 = 4320
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Which two triangles are congruent by ASA? <br> MP bisects QO and angle MQP congruent angle ROP
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Step-by-step explanation:

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3 years ago
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What is Radical 84/Two radical three
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Radical 7

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3 years ago
A random sample of 625 10-ounce cans of fruit nectar is drawn from among all cans produced in a run. Prior experience has shown
diamong [38]

Answer:

10.57% probability that the mean contents of the 625 sample cans is less than 9.995 ounces.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 10, \sigma = 0.1, n = 625, s = \frac{0.1}{\sqrt{625}} = 0.004

What is the probability that the mean contents of the 625 sample cans is less than 9.995 ounces?

This is the pvalue of Z when X = 9.995. So

Z = \frac{X - \mu}{s}

Z = \frac{9.995 - 10}{0.004}

Z = -1.25

Z = -1.25 has a pvalue of 0.1057

So there is a 10.57% probability that the mean contents of the 625 sample cans is less than 9.995 ounces.

5 0
2 years ago
a jar contains 30 red, 40 blue, and 50 white buttons. you pick one button at random, find the probability that is not white
Jet001 [13]

Answer:

<em>The probability that the button is not white is </em><em>0.583</em>

Step-by-step explanation:

The jar contains 30 red, 40 blue, and 50 white buttons.

In total there are 120 buttons in the jar. So,

|\ S\ |=120

Let us assume that, A be the event of picking white buttons, so

|\ A\ |=50

So the probability of picking white button is,

P(A)=\dfrac{|\ A\ |}{|\ S\ |}=\dfrac{50}{120}

Then the event A^c will be not picking up white buttons, so the probability of not picking up white buttons is,

P(A^c)=1-P(A)=1-\dfrac{50}{120}=\dfrac{70}{120}=\dfrac{7}{12}=0.583

3 0
3 years ago
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