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3 years ago

Is it possible for 0 and a positive number to have a common multiple?

Mathematics

2 answers:

luda_lava [24]3 years ago

5 0

The answer is yes. I hope I helped you !good luck to the home work

marin [14]3 years ago

3 0

We don't consider 0 a common multiple. So I don't think it is.

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(Fill in the blank)<br> Solve the following equation.

swat32

Answer:

there is no ecuashon

Step-by-step explanation:

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3 years ago

Read 2 more answers

A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pCI/L of radon. The resulting reading

valkas [14]

Answer:

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu \neq 100$

$t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921$

$p_v =2*P(t_{11}$

Step-by-step explanation:

1) Data given and notation

Data: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4

We can calculate the sample mean and deviation for this data with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}$

The results obtained are:

$\bar X=98.375$ represent the sample mean

$s=0.6.109$ represent the sample standard deviation

$n=12$ sample size

$\mu_o =100$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

2) State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is equal to 100pCL/L, the system of hypothesis are :

Null hypothesis: $\mu = 100$

Alternative hypothesis: $\mu \neq 100$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

3) Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921$

4) P-value

First we need to find the degrees of freedom for the statistic given by:

$df=n-1=12-1=11$

Since is a two sided test the p value would given by:

$p_v =2*P(t_{11}$

5) Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significant different from 100 at 5% of significance.

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3 years ago

Tara had $350 in her bank account she received three birthday checks for $35 each and but bought two new video games for $41 eac

goldfiish [28.3K]

We conclude that the final amount that she has in her bank account is $375.

<h3>How much has Tara in her bank account now?</h3>

Here we need to perform some additions and subtractions. We know that Tara starts with $350 in her account.

Then she received 3 checks for $35 each, so at this point she has:

$350 + 3*$35 = $455

Now she buys 2 video games for $41 each, so we need to subtract two times $41, we will get:

$455 - 2*$41 = $375

We conclude that the final amount that she has in her bank account is $375.

If you want to learn more about additions and subtractions:

brainly.com/question/25421984

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1 year ago

A survey of 85 families showed that 36 owned at least one DVD player. Find the 99% confidence interval estimate of the true prop

Katyanochek1 [597]

Answer: $(0.367,\ 0.473)$

Step-by-step explanation:

The confidence interval for population mean is given by :-

$\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where $\hat{p}$ is the sample proportion, n is the sample size , $z_{\alpha/2}$ is the critical z-value.

Given : Significance level : $\alpha:1-0.99=0.01$

Sample size : n= 85

Critical value : $z_{\alpha/2}=2.576$

Sample proportion: $\hat{p}=\dfrac{36}{85}\approx0.42$

Now, the 99% confidence level will be :

$\hat{p}\pmz_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}\\\\=0.42\pm(2.576)\sqrt{\dfrac{0.42(1-0.42)}{85}}\\\\\approx0.42\pm0.053\\\\=(0.42-0.053,\ 0.42+0.053)=(0.367,\ 0.473)$

Hence, the 99% confidence interval estimate of the true proportion of families who own at least one DVD player is $(0.367,\ 0.473)$

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3 years ago

Jillian wants to download a new ringtone to her cell phone. There are 954 music tones, 158 voice tones, and 63 sound effects to

guapka [62]

Answer:

Step-by-step explanation:

Just take all three numbers (954, 158, 63) then add them all up.

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2 years ago