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2 years ago

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A total of $12000 was invested in two types of bonds. One pays 8% simple interest while the other pays 10.5%. Last year, the ann

ual interest earned on the two investments was $1,145. How much was invested at each rate?

1 answer:

prisoha [69]2 years ago

7 0

Answer:

thanks for the points

Step-by-step explanation:

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Hard times In June 2010, a random poll of 800 working men found that 9% had taken on a second job to help pay the bills. (www.ca

Phoenix [80]

Answer:

a) The 95% confidence interval would be given (0.0702;0.1098).

We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)

b) Since the confidence interval NOT contains the value 0.06 we have anough evidence to reject the claim at 5% of significance.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$p$ represent the real population proportion of interest

$\hat p =0.09$ represent the estimated proportion for the sample

n=800 is the sample size required

$z$ represent the critical value for the margin of error

Confidence =0.95 or 95%

Part a

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

The margin of error is given by :

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

$Me=1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0198$

And replacing into the confidence interval formula we got:

$0.09 - 1.96 \sqrt{\frac{0.09(1-0.09)}{800}}=0.0702$

$0.09 + 1.96 \sqrt{\frac{0.108(1-0.09)}{800}}=0.1098$

And the 95% confidence interval would be given (0.0702;0.1098).

We can conclude that the true population proportion at 95% of confidence is between (0.0702;0.1098)

Part b

Since the confidence interval NOT contains the value 0.06 we have anough evidence to reject the claim at 5% of significance.

6 0

2 years ago

The height, h, of a falling object t seconds after it is dropped from a platform 300 feet above the ground is modeled by the fun

Bogdan [553]

Answer:

Average rate of fall = $\frac{h(3)-h(0)}{3}$

Step-by-step explanation:

Since they give you an expression for the position of the object as a function of time, one can derive the average velocity of the object during the first 3 seconds of the fall by estimating the position of the object at time 0, the position of the object at time 3 seconds, finding the displacement of the object (distance covered) in that interval of time (via a subtraction of these two quantities), and finally dividing this change by the elapsed time (3 seconds) thus using the concept that velocity is the quotient between distance covered over the elapsed time:

Average rate of fall = (h(3) - h(0))/ 3

Average rate of fall = $\frac{h(3)-h(0)}{3}$

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