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3 years ago

Which of the following situations could be solved using the equation 7h+1=29

Mathematics

1 answer:

Alina [70]3 years ago

5 0

Answer:

7h+1=29 is equal to 7 x 4 + 1=29

Step-by-step explanation:

We know that 7 times 4 = 28. So the value of h is 4. And of course, you add the one.

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Step-by-step explanation:

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4 years ago

Read 2 more answers

A manufacturer of paper coffee cups would like to estimate the proportion of cups that are defective (tears, broken seems, etc.)

Maksim231197 [3]

Answer:

The 95% confidence interval is $0.0503 < p < 0.1297$

The sample proportion is $\r p = 0.09$

The critical value is $Z_{\frac{\alpha }{2} } = 1.96$

The standard error is $SE =0.020$

Step-by-step explanation:

From the question we are told that

The sample size is n = 200

The number of defective is k = 18

The null hypothesis is $H_o : p = 0.08$

The alternative hypothesis is $H_a : p > 0.08$

Generally the sample proportion is mathematically evaluated as

$\r p = \frac{18}{200}$

$\r p = 0.09$

Given that the confidence level is 95% then the level of significance is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5\%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{ \alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the standard of error is mathematically represented as

$SE = \sqrt{\frac{\r p (1 - \r p)}{n} }$

substituting values

$SE = \sqrt{\frac{0.09 (1 - 0.09)}{200} }$

$SE =0.020$

The margin of error is

$E = Z_{\frac{ \alpha }{2} } * SE$

=> $E = 1.96 * 0.020$

=> $E = 0.0397$

The 95% confidence interval is mathematically represented as

$\r p - E < \mu < p < \r p + E$

=> $0.09 - 0.0397 < \mu < p < 0.09 + 0.0397$

=> $0.0503 < p < 0.1297$

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lisabon 2012 [21]

Step-by-step explanation:

The figure below shows a portion of the graph of the function $j\left(x\right) \ = \ 4^{x-2}$ , hence the average rate of change (slope of the blue line) between the $x$ and $x+h$ is

$\text{Average rate of change} \ = \ \displaystyle\frac{\Delta y}{\Delta x} \\ \\ \rule{3.7cm}{0cm} = \dsiplaystyle\frac{f\left(x+h\right) \ - \ f\left(x\right)}{\left(x \ + \ h \right) \ - \ x} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{f\left(x + h\right) \ - \ f\left(x\right)}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x+h-2} \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{4^{x-2+h} \ - \ 4^{x-2}}{h}$

$\\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h}\right) \ - \ 4^{x-2}}{h} \\ \\ \\ \rule{3.7cm}{0cm} = \displaystyle\frac{\left(4^{x-2}\right)\left(4^{h} \ - \ 1 \right)}{h}$

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2 years ago