Jose bought 6 flowers for his mom 2/3 of them where roses How many roses are there

A scientist measures a substance to be 0.6 grams. Calculate the percent of error in the measurement. Show all work for full cred

Anni [7]

A measurement of 0.6 grams means that the 'true' answer lies between
<span>0.55 and 0.65 </span>

In general, the uncertainty in a single measurement from a single instrument is half the least precise decimal given for the instrument.
The precision for this measurement is in the tenths place.
so we take 0.1 /2 = 0.05, and divide by the measurement

<span>0.05 / 0.6 = 0.08333 = 8.3 %</span>

6 0

3 years ago

A square has a side that measures 5.5 units. What is the area of a circle with a circumference that equals the perimeter of the

Alexeev081 [22]

Answer:

Answer choice A) About 38.47 square units

Step-by-step explanation:

Since all four sides of a square have the same length, the perimeter of a square is just 4 times one of the side lengths. The perimeter of the square and therefore the circumference of the circle is 5.5*4=22. The circumference of a circle is 2 times the radius multiplied by pi. The radius of this circle is therefore:

$\dfrac{22}{2\pi}\approx 3.503$

Since the area of a circle is $\pi r^2$ , the area of this circle is:

$\pi \cdot 3.503^2= 3.14\cdot 12.271\approx 38.47$

Hope this helps!

4 0

2 years ago

The width of Estelle's rectangular living room is 4 meters, and the distance between opposite corners is 7 meters. What is the l

Crank

Answer:

5.7 meters

Step-by-step explanation:

Image a rectangle being cut diagonally in half, that would lead to two triangles. We can use the pythagorean theorem to find the missing side.

7 is the hypotenuse and 4 is the side length

4^2+x^2=7^2

16+x^2=49

sqrx=sqr(49-16)

x=sqr33

x=5.74

Rounded to the nearest tenth is 5.7

6 0

2 years ago

Molly has a jar containing 4 yellow, 3 blue, 3 green, and 2 orange marbles

NeTakaya

Answer: .111 with a bar notation over the last 1 or 11.1%

Step-by-step explanation:

There are 12 marbles in total. There are 4 yellow marbles. Therefore there is a 4/12 probability of Randomly picking a Yellow marble the first time . Since she placed the marble back, there is a 4/12 probability of getting a yellow marble the 2nd time.

(4/12)(4/12) = 16/144 = 1/9 = .111 with a bar notation over the last 1 or 11.1% (about 11%)

6 0

2 years ago

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as t

skad [1K]

Answer:

a) 0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

b) 0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

c) 0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

d) None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The CPA Practice Advisor reports that the mean preparation fee for 2017 federal income tax returns was $273. Use this price as the population mean and assume the population standard deviation of preparation fees is $100.

This means that $\mu = 273, \sigma = 100$

A) What is the probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 30, s = \frac{100}{\sqrt{30}}$

The probability is the p-value of Z when X = 273 + 16 = 289 subtracted by the p-value of Z when X = 273 - 16 = 257. So

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{30}}}$

$Z = 0.88$

$Z = 0.88$ has a p-value of 0.8106

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{30}}}$

$Z = -0.88$

$Z = -0.88$ has a p-value of 0.1894

0.8106 - 0.1894 = 0.6212

0.6212 = 62.12% probability that the mean price for a sample of 30 federal income tax returns is within $16 of the population mean.

B) What is the probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 50, s = \frac{100}{\sqrt{50}}$

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{50}}}$

$Z = 1.13$

$Z = 1.13$ has a p-value of 0.8708

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{50}}}$

$Z = -1.13$

$Z = -1.13$ has a p-value of 0.1292

0.8708 - 0.1292 = 0.7416

0.7416 = 74.16% probability that the mean price for a sample of 50 federal income tax returns is within $16 of the population mean.

C) What is the probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean?

Sample of 30 means that $n = 100, s = \frac{100}{\sqrt{100}}$

X = 289

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{289 - 273}{\frac{100}{\sqrt{100}}}$

$Z = 1.6$

$Z = 1.6$ has a p-value of 0.9452

X = 257

$Z = \frac{X - \mu}{s}$

$Z = \frac{257 - 273}{\frac{100}{\sqrt{100}}}$

$Z = -1.6$

$Z = -1.6$ has a p-value of 0.0648

0.9452 - 0.0648 =

0.8804 = 88.04% probability that the mean price for a sample of 100 federal income tax returns is within $16 of the population mean.

D) Which, if any of the sample sizes in part (a), (b), and (c) would you recommend to ensure at least a .95 probability that the same mean is withing $16 of the population mean?

None of them ensure, that one which comes closer is a sample size of 100 in option c), to guarantee, we need to keep increasing the sample size.

6 0

2 years ago