Find the values of k for which y = kx + 1 is a tangent to the curve y = 2x^2 + x +3

One number is seven less than four times the other number. The sum of the two numbers is 93. Find the numbers.

ivann1987 [24]

You can write the two equations x = 4y - 7 and x + y = 93; substitute the value of x into the other equation

(4y - 7) + y = 93; combine like terms

5y - 7 = 93; add 7 to both sides

5y = 100; divide both sides by 5

y = 20; substitute this value into the other equation

x + 20 = 93

x = 73

So the numbers are 20 and 73

6 0

3 years ago

Of 575 broiler chickens purchased from various kinds of food stores in different regions of a country and tested for types of ba

chubhunter [2.5K]

Answer:

a) $0.68 - 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.630$

$0.68 + 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.730$

And the 99% confidence interval would be given (0.630;0.730).

b) We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)

c) For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval are satisfied, so then the results can be assumed for all the population

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

$\hat p =0.68$ represent the estimated proportion

n=575 is the sample size required (variable of interest)

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

The population proportion have the following distribution

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

Part a

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 99% confidence interval the value of $\alpha=1-0.99=0.01$ and $\alpha/2=0.005$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=2.58$

And replacing into the confidence interval formula we got:

$0.68 - 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.630$

$0.68 + 2.58 \sqrt{\frac{0.68(1-0.68)}{575}}=0.730$

And the 99% confidence interval would be given (0.630;0.730).

Part b

We are 99% confident that the true proportion of infected chickens are between 0.63 (63%) and 0.73 (73%)

Part c

For this case the answer would be No. Since all the required assumptions in order to construct the confidence interval are satisfied, so then the results can be assumed for all the population

4 0

3 years ago

PLS I NEED HELP ANSWER WILL GIVE BRAINLIEST ANSWER BOTH QUESTIONS ONE AND TWO

KiRa [710]

I hope this is helpful