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3 years ago

A number pattern starts with 10 and follows the rule multiply by 3 What is true

Mathematics

1 answer:

lesya692 [45]3 years ago

3 0

Answer:

B) They have a 0 in the ones place.

Step-by-step explanation:

Given

Number pattern that starts with 10.

Rule: Multiply by 3

Required

Select which statement is truth of the pattern.

First, the pattern needs to be listed out.

Given that the pattern starts from 10 and progresses by multiplying current term by 3.

The pattern is as follows;

10, 30, 90, 270, 810, 2430, 7290, .....

By comparing the given options with the above pattern of numbers. We have:

A) They are odd numbers.

This statement is false because the above pattern of numbers are all even and can never be odd

B) They have a 0 in the ones place.

This statement is truth because all the numbers in the above progression ends in 0. This 0 represents the unit of one's place.

Hence, this statement is truth.

C) They have a 3 in the tens place.

This statement is false;

This is so because they have different digits at tens place. The tens place is the digit prior to the last digit of each number.

D) They can be odd or ever numbers.

As pointed out in (A) above, the number can only be even number because the progression starts with an even number (10) and irrespective of what it's been multiplied by, it'll always remain an even number.

Hence,

Only option B is correct/truth.

Others are incorrect

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The diameters of bearings used in an aircraft landing gear assembly have a standard deviation of ???? = 0.0020 cm. A random samp

vovikov84 [41]

Answer:

(a) We conclude after testing that mean diameter is 8.2500 cm.

(b) P-value of test = 2 x 0.0005% = 1 x $10^{-5}$ .

(c) 95% confidence interval on the mean diameter = [8.2525 , 8.2545]

Step-by-step explanation:

We are given with the population standard deviation, $\sigma$ = 0.0020 cm

Sample Mean, Xbar = 8.2535 cm and Sample size, n = 15

(a) Let Null Hypothesis, $H_0$ : Mean Diameter, $\mu$ = 8.2500 cm

Alternate Hypothesis, $H_1$ : Mean Diameter, $\mu$ $\neq$ 8.2500 cm{Given two sided}

So, Test Statistics for testing this hypothesis is given by;

$\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ follows Standard Normal distribution

After putting each value, Test Statistics = $\frac{8.2535-8.2500}{\frac{0.0020}{\sqrt{15} } }$ = 6.778

Now we are given with the level of significance of 5% and at this level of significance our z score has a value of 1.96 as it is two sided alternative.

<em>Since our test statistics does not lie in the rejection region{value smaller than 1.96} as 6.778>1.96 so we have sufficient evidence to accept null hypothesis and conclude that the mean diameter is 8.2500 cm.</em>

(b) P-value is the exact % where test statistics lie.

For calculating P-value , our test statistics has a value of 6.778

So, P(Z > 6.778) = Since in the Z table the highest value for test statistics is given as 4.4172 and our test statistics has value higher than this so we conclude that P - value is smaller than 2 x 0.0005% { Here 2 is multiplied with the % value of 4.4172 because of two sided alternative hypothesis}

Hence P-value of test = 2 x 0.0005% = 1 x $10^{-5}$ .

Probability(-1.96 < N(0,1) < 1.96) = 0.95 { This indicates that at 5% level of significance our Z score will lie between area of -1.96 to 1.96.

P(-1.96 < $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96\frac{\sigma}{\sqrt{n} }$ < $Xbar - \mu$ < $1.96\frac{\sigma}{\sqrt{n} }$ ) = 0.95