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KIM [24]
4 years ago
7

compare the two numbers to find which is greater. explain how you can compare them without writing them in standard notation fir

st. 4.5×10^6 2.1×10^8
Mathematics
1 answer:
Anna71 [15]4 years ago
5 0

The number with the largest exponent is the largest value overall. In this case, that would be 2.1 x 10^8 which is the largest value since 8 is the largest exponent.

Note: The number 4.5 x 10^6 is the number 4.5 million, while the number 2.1 x 10^8 is the number 210 million. The positive exponent means "move the decimal point that number of times to the right". So 4.5 x 10^6 means "move the decimal point 6 spots to the right" so you can convert to standard form. The larger the exponent, the more places the decimal point moves to the right.

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Cost of a CD: $17.50<br> Markup: 45%
Marizza181 [45]

Answer:

25.375

Step-by-step explanation:

17.50×45%=7.875

17.50+7.875=25.375 new price

8 0
2 years ago
93=30-e
iren [92.7K]

Answer:

93+30=123 so e equals 123

Step-by-step explanation:

To check, do 123-30=93

5 0
3 years ago
HELP I NEED HELP ASAP
Allisa [31]

Answer:

3x^2-6x+3=0

x^2-2x+1=0

x^2-x-x+1=0

x(x-1)-(x-1)=0

(x-1)(x-1)=0

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5 0
3 years ago
What is the perimeter of the regular hexagon ? <br> 3m
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7 0
4 years ago
A random sample of n = 45 observations from a quantitative population produced a mean x = 2.5 and a standard deviation s = 0.26.
oee [108]

Answer:

P-value (t=2.58) = 0.0066.

Note: as we are using the sample standard deviation, a t-statistic is appropiate instead os a z-statistic.

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the population mean μ exceeds 2.4.

Then, the null and alternative hypothesis are:

H_0: \mu=2.4\\\\H_a:\mu> 2.4

The significance level is 0.05.

The sample has a size n=45.

The sample mean is M=2.5.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.26.

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

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.26}{\sqrt{45}}=0.0388

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{2.5-2.4}{0.0388}=\dfrac{0.1}{0.0388}=2.58

The degrees of freedom for this sample size are:

df=n-1=45-1=44

This test is a right-tailed test, with 44 degrees of freedom and t=2.58, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>2.5801)=0.0066

As the P-value (0.0066) is smaller than the significance level (0.05), the effect is  significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the population mean μ exceeds 2.4.

7 0
3 years ago
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