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Ket [755]
4 years ago
10

What is -7x3(4+-5)=​

Mathematics
1 answer:
nlexa [21]4 years ago
6 0

\text{Hey there!}

\bf{-7\times(4+(-5))}

\bf{First, you\ need\ to\ do \ is\ PEMDAS}\\\\\\\text{Parentheses, Exponents, Multiplication, Division,Addition\ \& Subtraction}

\bf{ -7\times3(4+(-5))}\\\bf{-7\times3=-21}\\\bf{-21(4+(-5))}\\\bf{4+(-5)= -1}\\\bf{-21\times-1=21}\\\\\boxed{\boxed{\bf{Answer: 21}}}\checkmark

\text{Good luck on your assignment and enjoy your day!}\\\\\\\frak{LoveYourselfFirst:)}

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Step-by-step explanation:

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3 years ago
Of 1000 randomly selected cases of lung cancer, 823 resulted in death within 10 years.
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Answer:

a) 0.823 - 1.96\sqrt{\frac{0.823(1-0.823)}{1000}}=0.799

0.823 + 1.96\sqrt{\frac{0.823(1-0.823)}{1000}}=0.847

The 95% confidence interval would be given by (0.799;0.847)

b) n=\frac{0.823(1-0.823)}{(\frac{0.03}{1.96})^2}=621.79  

And rounded up we have that n=622

c) n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

And rounded up we have that n=1068

Step-by-step explanation:

Part a

\hat p=\frac{823}{1000}=0.823

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 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The confidence interval for the mean is given by the following formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

If we replace the values obtained we got:

0.823 - 1.96\sqrt{\frac{0.823(1-0.823)}{1000}}=0.799

0.823 + 1.96\sqrt{\frac{0.823(1-0.823)}{1000}}=0.847

The 95% confidence interval would be given by (0.799;0.847)

Part b

The margin of error for the proportion interval is given by this formula:  

ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}    (a)  

And on this case we have that ME =\pm 0.03 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

And replacing into equation (b) the values from part a we got:

n=\frac{0.823(1-0.823)}{(\frac{0.03}{1.96})^2}=621.79  

And rounded up we have that n=622

Part c

n=\frac{0.5(1-0.5)}{(\frac{0.03}{1.96})^2}=1067.11  

And rounded up we have that n=1068

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