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fomenos
3 years ago
8

James is getting ready for wrestling season. As part of his preparation, he plans to lose 5% of his body weight James currently

weighs 100 pounds How much will he weigh, in pounds, after he loses 5% of his weight​
Mathematics
2 answers:
chubhunter [2.5K]3 years ago
3 0

Multiply 100 pounds by 5%: 100 x 0.05 = 5 pounds.

Subtract that from the original weight:

100 - 5 = 95

He will weigh 95 pounds

DanielleElmas [232]3 years ago
3 0

Answer:

95 pounds

Step-by-step explanation:

5% as a decimal is 0.05 (because it is written as five one hundredths)

0.05x100=5  this represents how much he will lose

100-5=95

he will weigh 95 pounds

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Step-by-step explanation:

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Solve only if you know the solution and show work.
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\displaystyle\int\frac{\cos x+3\sin x+7}{\cos x+\sin x+1}\,\mathrm dx=\int\mathrm dx+2\int\frac{\sin x+3}{\cos x+\sin x+1}\,\mathrm dx

For the remaining integral, let t=\tan\dfrac x2. Then

\sin x=\sin\left(2\times\dfrac x2\right)=2\sin\dfrac x2\cos\dfrac x2=\dfrac{2t}{1+t^2}
\cos x=\cos\left(2\times\dfrac x2\right)=\cos^2\dfrac x2-\sin^2\dfrac x2=\dfrac{1-t^2}{1+t^2}

and

\mathrm dt=\dfrac12\sec^2\dfrac x2\,\mathrm dx\implies \mathrm dx=2\cos^2\dfrac x2\,\mathrm dt=\dfrac2{1+t^2}\,\mathrm dt

Now the integral is

\displaystyle\int\mathrm dx+2\int\frac{\dfrac{2t}{1+t^2}+3}{\dfrac{1-t^2}{1+t^2}+\dfrac{2t}{1+t^2}+1}\times\frac2{1+t^2}\,\mathrm dt

The first integral is trivial, so we'll focus on the latter one. You have

\displaystyle2\int\frac{2t+3(1+t^2)}{(1-t^2+2t+1+t^2)(1+t^2)}\,\mathrm dt=2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt

Decompose the integrand into partial fractions:

\dfrac{3t^2+2t+3}{(1+t)(1+t^2)}=\dfrac2{1+t}+\dfrac{1+t}{1+t^2}

so you have

\displaystyle2\int\frac{3t^2+2t+3}{(1+t)(1+t^2)}\,\mathrm dt=4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt

which are all standard integrals. You end up with

\displaystyle\int\mathrm dx+4\int\frac{\mathrm dt}{1+t}+2\int\frac{\mathrm dt}{1+t^2}+\int\frac{2t}{1+t^2}\,\mathrm dt
=x+4\ln|1+t|+2\arctan t+\ln(1+t^2)+C
=x+4\ln\left|1+\tan\dfrac x2\right|+2\arctan\left(\arctan\dfrac x2\right)+\ln\left(1+\tan^2\dfrac x2\right)+C
=2x+4\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)+C

To try to get the terms to match up with the available answers, let's add and subtract \ln\left|1+\tan\dfrac x2\right| to get

2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left(\sec^2\dfrac x2\right)-\ln\left|1+\tan\dfrac x2\right|+C
2x+5\ln\left|1+\tan\dfrac x2\right|+\ln\left|\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}\right|+C

which suggests A may be the answer. To make sure this is the case, show that

\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\sin x+\cos x+1

You have

\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\cos^2\dfrac x2+\sin\dfrac x2\cos\dfrac x2}
\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac1{\dfrac{1+\cos x}2+\dfrac{\sin x}2}
\dfrac{\sec^2\dfrac x2}{1+\tan\dfrac x2}=\dfrac2{\cos x+\sin x+1}

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How to do this question plz ​
Akimi4 [234]

Answer:

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Step-by-step explanation:

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8 0
3 years ago
A random sample of 11 nursing students from Group 1 resulted in a mean score of 41.3 with a standard deviation of 6.8. A random
Lina20 [59]

Answer:

At 1% significance level,  this difference is considered to be extremely statistically significant.

Step-by-step explanation:

 Group   Group One     Group Two  

Mean 41.300 54.800

SD 6.800 6.000

SEM 2.050 1.604

N 11      14      

H0: Mean of group I = Mean of group II

Ha: Mean of group I < mean of group II

(Left tailed test at 1% significance level)

  The mean of Group One minus Group Two equals -13.500

 standard error of difference = 2.563

 t = 5.2681

 df = 23

 p value= 0.00005

Since p < significance level, reject H0

6 0
3 years ago
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