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lozanna [386]
3 years ago
13

Jackson is playing a video game that has 4 castles to conquer within each level. Jackson has completed 6 levels so far. If x rep

resents the number of remaining levels to complete, which of the following equations can be used to find the total number of castles Jackson will have conquered once he completes the final level? A. y = 4x + 24 B. y = 6x + 4 C. y = 4x + 6 D. y = 24x + 4
Mathematics
1 answer:
Alenkinab [10]3 years ago
8 0
Each level has 4 castles to conquered. Jackson completed 6.
Let x be the number of remaining levels and y is the total number of castles to defeat

For every x, there are 4 castles to conquered, 4x represents number of remaining levels and corresponding castles.

Since he has already conquered 6 levels 4(6) = 24 so
y=4x+24, A.
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Step-by-step explanation:

if a dairy farm has 630 cows and the same number of cows are placed in each pasture the total number of cows in each pasture is 630

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Step-by-step explanation:

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Simplify x+2/x^2+2x-3 divide by x+2/x^2-x
ira [324]

\dfrac{\frac{x+2}{x^2+2x-3}}{\frac{x+2}{x^2-x}}

If x\neq-2, then we can immediately cancel the factors of x+2:

\dfrac{\frac1{x^2+2x-3}}{\frac1{x^2-x}}=\dfrac{x^2-x}{x^2+2x-3}

Factorize the numerator and denominator:

x^2-x=x(x-1)

x^2+2x-3=(x+3)(x-1)

Next, if x\neq1, then

\dfrac{x^2-x}{x^2+2x-3}=\dfrac{x(x-1)}{(x+3)(x-1)}=\dfrac x{x+3}

8 0
3 years ago
Your state is considering raising the legal age for consumption of alcoholic beverages to 21 years old. How large a sample size
julsineya [31]

Answer:

n=\frac{0.5(1-0.5)}{(\frac{0.01}{2.58})^2}=16641  

And rounded up we have that n=16641

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".  

The population proportion have the following distribution

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

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by \alpha=1-0.99=0.01 and \alpha/2 =0.005. And the critical value would be given by:

z_{\alpha/2}=-2.58, t_{1-\alpha/2}=2.58

The margin of error for the proportion interval is given by this formula:  

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

And on this case we have that ME =\pm 0.01 and we are interested in order to find the value of n, if we solve n from equation (a) we got:  

n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}   (b)  

For this case we can use as estimator for the proportion \hat p =0.5, since we don't have any other previous info. And replacing into equation (b) the values from part a we got:

n=\frac{0.5(1-0.5)}{(\frac{0.01}{2.58})^2}=16641  

And rounded up we have that n=16641

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igomit [66]

Answer:

qualitative

Step-by-step explanation:

bcos the question is in quality format

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