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3 years ago

For one dozen cookies you need to use of a cup of sugar,

Mathematics

1 answer:

Andre45 [30]3 years ago

6 0

Answer:

4 dozen

Step-by-step explanation:

You might be interested in

X^2+ 3 has a value of 28. What term in the sequence has a value of 28?

VikaD [51]

Answer:

5th term

Step-by-step explanation:

x^2 +3 = 28

Subtract 3 from each side

x^2+3-3 =28-3

x^2 = 25

Take the square root of each side

sqrt(x^2) = sqrt(25)

x = 5

(x could be -5, but there are not usually negative terms in a sequence)

5 0

3 years ago

Read 2 more answers

8x-6y=54 in slope intercept form please!!

NikAS [45]

Answer: y= 4/3x-9

Step-by-step explanation:

4 0

3 years ago

- 7(2-1) = 28<br> The solution is?

Aneli [31]

Answer:

Is there a x somewhere or is that litteraly how the equation is?

If so, -7 multiplied by 2 is -14,

-1 times -7 is 7,

add it together and you get

-7=28,

you could add 7 to both sides or subtract 28 from both sides and you'll get either 0=35 or -35=0

which would make the solution false since both sides are not the same

6 0

3 years ago

There are 2,000 eligible voters in a precinct. A total of 500 voters are randomly selected and asked whether they plan to vote f

Ann [662]

Answer:

$0.7 - 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.647$

$0.7 + 2.58 \sqrt{\frac{0.7(1-0.7)}{500}}=0.753$

And the 99% confidence interval would be given (0.647;0.753).

So the correct answer would be:

a. 0.647 and 0.753

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

The estimated population proportion for this case is:

$\hat p = \frac{350}{500}=0.7$

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.