Use pictures or words to explain how to use the sum of 13 to show how to add in any order

a red die is tossed and then a green die is tossed. what is the probability that the red die shows a six and the green die shows

vaieri [72.5K]

Assume that each of the 6 numbers has an equal probability of showing up for each die.

The probability of obtaining a six from the red die is 1/6, or
P(6 from the red die) = 1/6.
Similarly,
P(6 from the green die) = 1/6.

The tossing of the red die, followed by the tossing of the green die are independent events. Therefore
P(6 from the red die AND 6 from the green die) = (1/6)*(1/6) = 1/36.

Answer: 1/36

5 0

3 years ago

Find the slope of the line. -5y = -33 - 16

nata0808 [166]

It’s -33 and the y intercept is -5y

7 0

3 years ago

Read 2 more answers

How would I answer 1.75 times 6

Yuki888 [10]

Answer:10.5

Step-by-step explanation:You just add 1.75 6 times.

4 0

3 years ago

Read 2 more answers

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

Show All Work

Angelina_Jolie [31]

The unit rate you're trying to find is pages per day(or p/d), so the equation needs to have both a unit for pages and for days.
The equation we have is:
$5,249 = 29d$
If they read 5,249 pages, then we can include the unit for pages in the equation.
Since we also know that d is the number of days it took, you can replace d with days.
The equation becomes:
$5,249\ pages=29*(x\days)$
Now that we have one variable, we can solve for p/d:
$\frac{5,249\ pages}{29}=x\ days\\\\181 = x$
Thus it took them 181 days to read it all.

If Johnny read 5,249 pages over 181 days and the unit rate is pages per day(p/d), then the equation for finding p/d is:
$\frac{p\ pages}{d\ days}=\frac{5,249\ pages}{181\ days}=\frac{29\ pages}{1\ day}\\\\29\ \frac{pages}{day}$
Johnny read 29 pages per day.

8 0

3 years ago