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4 years ago

Work out 11.5% of 320

Mathematics

1 answer:

KiRa [710]4 years ago

7 0

11.5/100 (320)
184/5 (decimal:36.8)

hope this helps!

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Brenda bought five hats. A week later half

VashaNatasha [74]

Answer:

Step-by-step explanation:

23 divided by X. X=half of 23. but really, you need to add to it. other than that i can't help ya, all i know is that half of 23 is 11.5

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3 years ago

(-3, 2) (-4, 5)

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pls double check these im not super sure

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Step-by-step explanation:

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4 years ago

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Given P space equals space P subscript 0 space plus space rho g h, your objective is to determine the uncertainty of P by measur

son4ous [18]

Answer:

the g's contributing term for the overall uncertainty of P is $dP_g = [\frac{dg}{g}]$

Step-by-step explanation:

From the question we are told that

The pressure is $P = P_o + \rho gh$

The first step in determining the uncertainty of P in by obtaining the terms in the equation contributing to it uncertainty and to do that we take the Ln of both sides of the equation

$ln P = lnP_o + ln(\rho gh )$

=> $ln P = lnP_o + ln \rho + ln g + ln h$

Then the next step is to differentiate both sides of the equation

$\frac{d(ln P)}{dP} = \frac{d(ln P_o)}{dP_o} + \frac{d(ln \rho)}{d\rho} +\frac{d(ln g)}{dg} + \frac{d(ln h)}{dh}$

=> $\frac{dP}{P} = \frac{dP_o}{P_o} + \frac{d \rho}{\rho} +\frac{d g}{g} + \frac{d h}{h}$

We asked to obtain the contribution of the term g to the uncertainty of P

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4 years ago

. According to Management Accounting, salary figures for certified management accountants (CMAs) who are in the field less than

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Answer:

The Mean is not equal to $31,129.

Step-by-step explanation:

Consider the provided information.

The mean is $31129

Thus, the null hypothesis $H_0: \mu = 31129$

Alternative hypothesis $H_a: \mu \neq 31129$

A random sample of 15 firstyear CMAs in Denver produces a mean salary of $32,279, with a standard deviation of $1,797.

Thus the value mean of sample is 32279, standard deviation is 1797 and number of samples are 15.

Now use the 2 sided t-test.

$t=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

Now substitute the respective values in the above formula.

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$t=\frac{1150}{464}$

Test Value = 2.4785 approximately

Now, find the corresponding p-value in your t-table with DF(degree of freedom) 14.

p = 0.0265

as the value of p < 0.05, so you reject null hypothesis.

Thus, the Mean is not equal to $31,129.

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3 years ago

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It’s either -538 or +538 depends the circumstances

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3 years ago

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