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3 years ago

1/3 is part of which set of numbers?

Mathematics

2 answers:

nexus9112 [7]3 years ago

8 0

Answer:

$\frac{1}{3}$ Rational number as denominator is not equal to zero and numerator is a integer.

Scilla [17]3 years ago

7 0

Rational numbers. denoted by $\mathbb Q$

1/3 is clearly not a natural number or integer.

it is a fraction, =0.333 , it fits the definition of rational number ( $\frac pq$ ).

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2 years ago

suppose that the current canadian dollar to US dollar exchange rate is 0.80 CAD = $1 US and that the U.S. dollar price of an App

Deffense [45]

Answer: 256 CAD

Step-by-step explanation:

Hi, to answer this question we simply have to multiply the price in US dollars of the iphone (320) and multiply it by 0.80 ( Canadian dollar to US dollar exchange rate)

Mathematically speaking:

Since:

$1 US = 0.80 CAD

$320 x0.80 = 256 CAD

Feel free to ask for more if needed or if you did not understand something.

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3 years ago

An older person is 4 years older than 7 times the age of a younger person. The sum of their ages is 44. Find their ages

Fynjy0 [20]

Answer:

5 yr and 39 yr

Step-by-step explanation:

Let x = the age of the younger

then 7x + 4 = the age of the older

Now x + 7x + 4 = 44

8x + 4 = 44

8x = 40

x = 5

7x + 4 = 7(5) + 4 = 35 + 4 = 39

Check: 5 + 39 = 44

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3 years ago

How pure is 16-karat gold?

natta225 [31]

Answer:

here

Step-by-step explanation:

16 karat gold is 16 parts gold to 8 parts other material for a pure gold percentage of 67%.

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3 years ago

Read 2 more answers

A survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries,

Andru [333]

Answer:

A. Null and alternative hypothesis:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

B. Yes. At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

P-value = 0.00002

C. As the difference is calculated as (population 1 − population 2), being population 1: groceries and population 2: dinning out, and knowing there is evidence that the true mean difference is positive, we can say that the groceries annual credit card charge is higher than dinning out annual credit card charge.

The point estimate is the sample mean difference d=$840.

The 95% confidence interval for the mean difference between the population means is (490, 1190).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

Then, the null and alternative hypothesis are:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

The significance level is 0.05.

The sample has a size n=42.

The sample mean is M=840.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1123.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{1123}{\sqrt{42}}=173.2827$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{840-0}{173.2827}=\dfrac{840}{173.2827}=4.848$

The degrees of freedom for this sample size are:

$df=n-1=42-1=41$

This test is a two-tailed test, with 41 degrees of freedom and t=4.848, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>4.848)=0.00002$

As the P-value (0.00002) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

We have to calculate a 95% confidence interval for the mean difference.

The t-value for a 95% confidence interval and 41 degrees of freedom is t=2.02.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.02 \cdot 173.283=350$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 840-350=490\\\\UL=M+t \cdot s_M = 840+350=1190$

The 95% confidence interval for the mean difference is (490, 1190).

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3 years ago