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4 years ago

I wasnt paying attention

Mathematics

2 answers:

labwork [276]4 years ago

7 0

a equals 16. You multiply both sides be -10 and then you add 6.

ExtremeBDS [4]4 years ago

5 0

Hello!

First of all, let's flip our equation, just for proper formatting.

$\frac{a-6}{-10}$ =-1

First of all, let's multiply both sides of our equation by -10.

a-6=10

Now we add 6 to both sides.

a=16

I hope this helps!

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If a truck averages 12 miles to the gallon, how many pounds of carbon dioxide are emitted into the atmosphere when the truck tra

77julia77 [94]

Answer: 278.30 pounds (if the gasoline does not contain ethanol)

317.12 pounds (if the gasoline does contain ethanol)

Step-by-step explanation:

If the truck averages 12 miles per gallon, then in 170 miles it consumes:

170mi/12mi = 14.17 gallons.

We know that for one gallon consumed, the carbon dioxide emitted into the atmosphere is 19.64 pounds. (assuming it does not contain ethanol)

Then for 14.17 gallons consumed, we have a emission of:

14.17*19.64 pounds = 278.30 pounds

I the gasoline contains ethanlol, for a gallon the emiision is around 22.38 pounds.

In this case the total emission is:

14.17*22.38 pounds = 317.12 pounds

3 0

4 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

4 years ago

How to change 17/8 to a proper fraction

Dafna1 [17]

8 can go into 17 twice with one left over so
the answer is 2 1/8

Hope this helps!!!

3 0

3 years ago

Read 2 more answers

Using the Taylor rule, if inflation is 1 percent, desired inflation is 2 percent, and output is 2 percentage points below potent

antoniya [11.8K]

Answer:

1.5%