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2 years ago

Write 0.00658 in standard form

Mathematics

1 answer:

Svetllana [295]2 years ago

5 0

Answer:

0.00658 = 6.58 * 10^-3

Step-by-step explanation:

The standard form is a way of writing numbers easily in powers of 10.

To write a number in standard form, we have to move the decimal point to the front of the first non zero number.

Depending on the position of the decimal point to the first non zero number, movement can be towards the right or towards the left. When we move towards the right, our power of 10 will be negative, when we move in the left direction, our power of 10 will be positive

In this question, we shall be moving towards the right. Thus, our power of 10 is negative. We shall be moving towards the right 3 three times

Thus our power would be 10^-3

Thus our standard form will be 6.58 * 10^-3

Kindly note that another pointer is, if the value of our number to be written in standard form is less than zero, then the standard form will come in negative powers of 10. If the value of the number is greater or equal to 1 at least, then then the standard form will be in positive powers of 10

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To solve this problem you must follow the proccedure shown below:

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2 years ago

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Answer:

Step-by-step explanation:

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Step-by-step explanation:

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"

When we conduct a multiple regression we want to know about the relationship between several independent or predictor variables and a dependent or criterion variable.

If we assume that we have $k$ independent variables and we have $j=1,\dots,j$ individuals, we can define the following formulas of variation:

$SS_{total}=\sum_{j=1}^n (y_j-\bar y)^2$

$SS_{regression}=SS_{model}=\sum_{j=1}^n (\hat y_{j}-\bar y)^2$

$SS_{error}=\sum_{j=1}^n (y_{j}-\hat y_j)^2$

And we have this property

$SST=SS_{regression}+SS_{error}$

The degrees of freedom for the model on this case is given by $df_{model}=df_{regression}=k=2$ where k =2 represent the number of variables.

The degrees of freedom for the error on this case is given by $df_{error}=N-k-1=24$ . Sinc we know k we can find N.

$N=24+k+1=24+2+1=27$

And the total degrees of freedom would be $df=N-1=27 -1 =26$

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2 years ago

Two brands of AAA batteries are tested in order to compare their voltage. The data summary can be found below. Find the 93% conf

kobusy [5.1K]

Answer:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}$

And replacing we got:

$(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903$

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And the confidence interval for the difference of means would be given by:

$0.2903 \leq \mu_1 -\mu_2 \leq 0.5097$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

We have the following data given:

$\bar X_1 = 9.2 , \bar X_2 = 8.8, \sigma_1= 0.3, n_1 = 27, \sigma_2 = 0.1, n_2 = 30$

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 93% of confidence, our significance level would be given by $\alpha=1-0.93=0.07$ and $\alpha/2 =0.035$ . And the critical value would be given by:

$z_{\alpha/2}=-1.811, z_{1-\alpha/2}=1.811$

The confidence interval is given by:

$(\bar X_1 -\bar X_2) \pm z_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_}{n_2}}$

And replacing we got:

$(9.2 -8.8) - 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.2903$

$(9.2 -8.8) + 1.811 \sqrt{\frac{0.3^2}{27}+\frac{0.1^2}{30}}= 0.5097$

And the confidence interval for the difference of means would be given by:

$0.2903 \leq \mu_1 -\mu_2 \leq 0.5097$

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2 years ago