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3 years ago

Which is not an advantage of using a histogram to represent data

Mathematics

1 answer:

Firdavs [7]3 years ago

8 0

When a histogram is used to represent data, the individual data values are not retained.

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2. The number of atoms of a particular chemical element is modelled by

____ [38]

9514 1404 393

Answer:

a) 600

b) see below

c) 1.26 hours

Step-by-step explanation:

a) The value of y when x=0 is the coefficient of the exponential term:

y = 600·3^(-0) = 600·1 = 600

There were 600 atoms to start.

b) see attached for a graph

c) The graph shows 150 atoms at t = 1.26, about 1.26 hours after the start of time counting.

If you want to find that value algebraically, substitute for y and solve for x. Logarithms are involved.

150 = 600·3^(-x)

150/600 = 3^(-x)

log(1/4) = -x·log(3)

x = -log(1/4)/log(3) = log(4)/log(3) ≈ 1.2618595

After about 1.26 hours, there were 150 atoms.

6 0

3 years ago

How many terms are there in the expression 2x - 5y + 3 + x?

Veseljchak [2.6K]

so, 3x-5y+3 is 3 terms

6 0

3 years ago

Read 2 more answers

A manufacturer produces crankshafts for an automobile engine. the crankshafts wear after 100,000 miles (0.0001 inch) is of inter

Daniel [21]

Part A:

Significant level:

α = 0.05

Null and alternative hypothesis:

h0 : μ = 3 vs h1: μ ≠ 3

Test statistics:

$z= \frac{\bar{x}-\mu}{\sigma/\sqrt{n}} \\ \\ = \frac{2.78-3}{0.9/\sqrt{15}} \\ \\ = \frac{-0.22}{0.2324} =-0.9467$

P-value:

P(-0.9467) = 0.1719

Since the test is a two-tailed test, p-value = 2(0.1719) = 0.3438

Conclusion:

Since the p-value is greater than the significant level, we fail to reject the null hypothesis and conclude that there is no sufficient evidence that the true mean is different from 3.

Part B:

The power of the test is given by:

$\beta=\phi\left(Z_{0.025}+ \frac{3-3.25}{0.9/\sqrt{15}}\right) -\phi\left(-Z_{0.025}+ \frac{3-3.25}{0.9/\sqrt{15}}\right) \\ \\ =\phi\left(1.96+ \frac{-0.25}{0.2324} \right)-\phi\left(-1.96+ \frac{-0.25}{0.2324} \right)=\phi(0.8842)-\phi(-3.0358) \\ \\ =0.8117-0.0012=0.8105$

Therefore, the power of the test if μ = 3.25 is 0.8105.

Part C:

The sample size that would be required to detect a true mean of 3.75 if we wanted the power to be at least 0.9 is obtained as follows:

$1-0.9=\phi\left(Z_{0.025}+ \frac{3-3.75}{0.9/\sqrt{n}}\right) -\phi\left(-Z_{0.025}+ \frac{3-3.75}{0.9/\sqrt{n}}\right) \\ \\ \Rightarrow0.1=\phi\left(1.96+ \frac{-0.75}{0.9/\sqrt{n}}\right)-\phi\left(-1.96+ \frac{-0.75}{0.9/\sqrt{n}} \right) \\ \\ =\phi\left(1.96+(-3.2415)\right)-\phi\left(1.96+(-3.2415)\right) \\ \\ \Rightarrow\frac{-0.75}{0.9/\sqrt{n}}=-3.2415 \\ \\ \Rightarrow\frac{0.9}{\sqrt{n}}=\frac{-0.75}{-3.2415}=0.2314 \\ \\ \Rightarrow\sqrt{n}=\frac{0.9}{0.2314}=3.8898$

$\Rightarrow n=(3.8898)^2=15.13$

Therefore, the sample size that would be required to detect a true mean of 3.75 if we wanted the power to be at least 0.9 is 16.

8 0

3 years ago

A two Way frequency table is shown below displaying the relationship between pizza type and crust choices for pizza. we took a s

makvit [3.9K]

There are 40 total Hawaiian Pizzas and 2 are thin crust.

The percentage = 2 /40 = 0.05 = 5%

8 0

3 years ago

Can someone give me the answers I don't get this at all

Vikentia [17]

5 to 8 right or there more

4 0

3 years ago

Read 2 more answers