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dexar [7]
1 year ago
9

What is the sum of the ten thousand digit and the million's digit of 1,234,567,890

Mathematics
1 answer:
zloy xaker [14]1 year ago
7 0

Answer: The ten thousand digit is 6.  And the millions digit is 4. So the answer is 10.

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2. Which expression is the exponential form of √30
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√30 
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Kevin buys a car. His car payment is $248 per month.After 55 payments how much was Kevin paying?
NNADVOKAT [17]

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hes paid 13,640

Step-by-step explanation:

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3 years ago
What is the approximate value of x in the equation below. log_3/4 25 = 3x-1
yuradex [85]

Answer: The approximate value of x is -3.396

Step-by-step explanation:

Given:  

Now,to find the value of x.

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We have,

Now,

Using value of:   and  we get,

on simplify:

Adding 1 both the sides, we get

Therefore, the approximate value of x for the equation    is  -3.396.

=

Step-by-step explanation:

5 0
2 years ago
We learned in that about 69.7% of 18-20 year olds consumed alcoholic beverages in 2008. We now consider a random sample of fifty
maw [93]

Answer:

(1) The expected number of people who would have consumed alcoholic beverages is 34.9.

(2) The standard deviation of people who would have consumed alcoholic beverages is 10.56.

(3) It is surprising that there were 45 or more people who have consumed alcoholic beverages.

Step-by-step explanation:

Let <em>X</em> = number of adults between 18 to 20 years consumed alcoholic beverages in 2008.

The probability of the random variable <em>X</em> is, <em>p</em> = 0.697.

A random sample of <em>n</em> = 50 adults in the age group 18 - 20 years is selected.

An adult, in the age group 18 - 20 years, consuming alcohol is independent of the others.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 50 and <em>p</em> = 0.697.

The probability mass function of a Binomial random variable <em>X</em> is:

P(X=x)={50\choose x}0.697^{x}(1-0.697)^{50-x};\ x=0,1,2,3...

(1)

Compute the expected value of <em>X</em> as follows:

E(X)=np\\=50\times 0.697\\=34.85\\\approx34.9

Thus, the expected number of people who would have consumed alcoholic beverages is 34.9.

(2)

Compute the standard deviation of <em>X</em> as follows:

SD(X)=\sqrt{np(1-p)}=\sqrt{50\times 0.697\times (1-0.697)}=10.55955\approx10.56

Thus, the standard deviation of people who would have consumed alcoholic beverages is 10.56.

(3)

Compute the probability of <em>X</em> ≥ 45 as follows:

P (<em>X</em> ≥ 45) = P (X = 45) + P (X = 46) + ... + P (X = 50)

                =\sum\limits^{50}_{x=45} {50\choose x}0.697^{x}(1-0.697)^{50-x}\\=0.0005+0.0001+0.00002+0.000003+0+0\\=0.000623\\\approx0.0006

The probability that 45 or more have consumed alcoholic beverages is 0.0006.

An unusual or surprising event is an event that has a very low probability of success, i.e. <em>p</em> < 0.05.

The probability of 45 or more have consumed alcoholic beverages is 0.0006. This probability value is very small.

Thus, it is surprising that there were 45 or more people who have consumed alcoholic beverages.

6 0
2 years ago
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