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Vikentia [17]
2 years ago
6

In math class, Sam correctly multiplied 3.625×10-8 by 500, then reported his product in scientific notation.

Mathematics
1 answer:
Minchanka [31]2 years ago
5 0
<h3><u>Question:</u></h3>

In math class, Sam correctly multiplied 3.625×10-8 by 500, then reported his product in scientific notation. What was the exponent in Sam’s product? A -16 B -10 C -7 D -5

<h3><u>Answer:</u></h3>

OPTION D

The exponent in Sam’s product is -5

<h3><u>Solution:</u></h3>

Given that,

In math class, Sam correctly multiplied 3.625 \times 10^{-8} by 500, then reported his product in scientific notation.

<em><u>To find: exponent in Sam’s product</u></em>

Let us first multiply 500 with 3.625 \times 10^{-8}

3.625 \times 10^{-8} \times 500 = 1812.5 \times 10^{-8}

Then Sam has reported this answer in scientific notation

Lets understand what is scientific notation of any number.

Scientific notation is a way of representing number between 1 and 10 incluing 1 and excluding 10 multiplied by appropriate power of 10.

Let us rewrite the Sam's answer in scientific notation

<em><u>The format for writing a number in scientific notation is:</u></em>

(first digit of the number) followed by (the decimal point) and then (all the rest of the digits of the number), times (10 to an appropriate power).

In 1812.5 \times 10^{-8} , we move the decimal 3 times to left side

Since the decimal is moved left side, we multiply by 10^3

1812.5 \times 10^{-8} = 1.8125 \times 10^3 \times 10^{-8} \\\\1812.5 \times 10^{-8} = 1.8125 \times 10^{-5}

Exponent is a quantity representing the power to which a given number or expression is to be raised

So in 1.8125 \times 10^{-5} exponent is -5

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b) If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

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A paired t-test is used to compare two population means where you have two samples in  which observations in one sample can be paired with observations in the other sample. For example  if we have Before-and-after observations (This problem) we can use it.  

The Q-Q plot, or quantile-quantile plot, "is a graphical tool to help us assess if a set of data plausibly came from some theoretical distribution such as a Normal or exponential".

Let put some notation  

x=value for A , y = value for B

A: 379, 378, 328, 372, 325, 304, 356, 309, 354, 318, 355, 392

B: 372, 376, 328, 368, 283, 252, 369, 321, 379, 303, 328, 411

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Are dependent since we are mesuring at the same individuals but on different times and with a different method

(b) Make a normal QQ plot of the differences within each pair. Is it reasonable to assume a normal population of differences?

The first step is calculate the difference d_i=A_i-B_i and we obtain this:

d: 7,2,0,4,42,52,-13,-12,-25,15,27,-19

In order to do the qqplot we can use the following R code:

d<-c(7,2,0,4,42,52,-13,-12,-25,15,27,-19)

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And the graph obtained is attached.

If we see the qq plot we don't have any significant deviation for the values and we don't have any heavy tail so we can conclude that we can approximate the differences with the normal distribution.

(c) Choose a test appropriate for the hypotheses above and justify your choice based on your answers to parts (a) and (b). Perform the test by computing a p-value, make a test decision, and state your conclusion in the context of the problem

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Alternative hypothesis: \mu_A -\mu_B \neq 0

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