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tangare [24]
2 years ago
9

There are 10 Superscript 9 bytes in a gigabyte. There are 10 Superscript 6 bytes in a megabyte. How many times greater is the st

orage capacity of a 1-gigabyte flash drive than a 1-megabyte flash drive? 3 times greater 10 times greater 1,000 times greater 3,000 times greater.
Mathematics
1 answer:
ohaa [14]2 years ago
8 0

To find the how many  times greater is the storage capacity of a 1-gigabyte flash drive than a 1-megabyte flash drive we have to apply exponent operation.

The storage capacity of a 1-gigabyte is 1000 greater than 1-megabyte flash drive.

Given:

The given exponent for gigabyte is,

10^9

The given exponent of megabyte is,

10^6

<h3>What is exponent rule?</h3>
  • The power rule of exponent rule is defined as if raise the a number with exponent to power then we will mulitply exponent to the power times.  
  • to change the negative exponent into a positive we have to invert the base.

Calculate the storage capacity.

\dfrac{10^9}{ 10^6}=10^9-6}\\&#10;\dfrac{10^9}{ 10^6}=10^3\\&#10;\dfrac{10^9}{ 10^6}=1000

Thus, the storage capacity of a 1-gigabyte is 1000 greater than 1-megabyte flash drive.

Learn more about exponent rule here:

brainly.com/question/2254060

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Answer:

(-4, -1)

Step-by-step explanation:

-3x - 8y = 20; y = 5x + 19

y = 5x + 19; -3x - 8y = 20

y = 5x + 19

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-3x - 8(5x + 19) = 20

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3 years ago
If P(x,y) is the point on the unit circle determined by real number 0, then tan0=​
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Step-by-step explanation:

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3 years ago
Test the null hypothesis Upper H 0 : (mu 1 minus mu 2 )equals 0H0: μ1−μ2=0 versus the alternative hypothesis Upper H Subscript a
Law Incorporation [45]

Answer:

The test statistic t is t=2.9037.

The null hypothesis is rejected.

For a significance level of 0.05, there is enough evidence to support the alternative hypothesis.

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>The sample 1, of size n1=25 has a mean of 1.15 and a standard deviation of 0.31. </em>

<em>The sample 2, of size n2=25 has a mean of 0.95 and a standard deviation of 0.15. </em>

This is a hypothesis test for the difference between populations means.

The null and alternative hypothesis are:

H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2\neq 0

The significance level is α=0.05.

The difference between sample means is Md=0.2.

M_d=M_1-M_2=1.15-0.95=0.2

The estimated standard error of the difference between means is computed using the formula:

s_{M_d}=\sqrt{\dfrac{\sigma_1^2+\sigma_2^2}{n}}=\sqrt{\dfrac{0.31^2+0.15^2}{25}}\\\\\\s_{M_d}=\sqrt{\dfrac{0.119}{25}}=\sqrt{0.005}=0.069

Then, we can calculate the t-statistic as:

t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{0.2-0}{0.069}=\dfrac{0.2}{0.069}=2.9037

The degrees of freedom for this test are:

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This test is a two-tailed test, with 48 degrees of freedom and t=2.9037, so the P-value for this test is calculated as (using a t-table):

P-value=2\cdot P(t>2.9037)=0.0056

As the P-value (0.0056) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

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The best answer to go with is c
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3 years ago
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