The sum of 1/11 and 1/15 is closest to....

2 answers:

4 0

$\frac{1}{11}+\frac{1}{15}=\frac{1\cdot15}{11\cdot15}+\frac{1\cdot11}{15\cdot11}=\frac{15}{165}+\frac{11}{165}=\frac{15+11}{165}=\frac{26}{165} < \frac{1}{2}\\\\\frac{1}{11}+\frac{1}{15}\approx0$

VMariaS [17]3 years ago

3 0

1/11=0.090
1/15=0.060
The sum of the two is 0.15

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An automobile manufacturer claims that its car has a 35.935.9 miles/gallon (MPG) rating. An independent testing firm has been co

Serjik [45]

<h2>Answer with explanation:</h2>

Let $\mu$ be the population mean.

For the given claim , we have

Null hypothesis : $H_0: \mu=35.9$

Alternative hypothesis : $H_a: \mu\neq35.9$

Since alternative hypothesis is two-tailed , so the test is a two-tailed test.

Given : Sample size : n=220 ;

Sample mean: $\overline{x}=35.6$ ;

Standard deviation: $s=2.2$

Test statistic for population mean:

$z=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}$

$z=\dfrac{35.6-35.9}{\dfrac{2.2}{\sqrt{220}}}\approx-2.02$

By using the standard normal distribution table of z , we have

P-value ( two tailed test ) : $2P(Z>|z|)=2(1-P(Z$

$=2(1-P(z$

Since , the P-value is greater than the significance level of $\alpha=0.02$ , it means we do not have sufficient evidence to reject the null hypothesis.