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3 years ago

Which is a reasonable estimate of the constant of variation

Mathematics

1 answer:

Leno4ka [110]3 years ago

7 0

Answer:

Direct Variation. Since k is constant we can find k when given any point by dividing the y-coordinate by the x-coordinate.

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a teacher plans a simulation to estimate the probability that a student will pick a vowel out of a bag of 26 tiles, each with a

Alex Ar [27]

Answer:

Yes

Step-by-step explanation:

Given that a teacher prepares 26 tiles with 5 vowels numbered 1 and 21 consonants numbered 2.

The probability for drawing vowel = $\frac{5}{26}$

Prob for consonant = $\frac{21}{26}$

If number of trials is atleast 30 we can expect reliable results.

Here the results are recorded for 120 times at random.

Since number of trials is large, we can expect a reliable and accurate results representing the actual probability.

This is because more the number of trials, the less would be the margin of error i.edeviationfrom the expected probability would be minimum

8 0

4 years ago

Read 2 more answers

Carbon-14 has a half-life of 5,730 years. If a sample contained 70 mg originally, how much is left after 17,190 years?

kow [346]

17,190 / 5,730 = 3. This means we must divide 70mg in half 3 times

70 / 2 = 35

35 / 2 = 17.5

17.5 / 2 = 8.75

There will be 8.75 mg left after 17,190 years.

7 0

3 years ago

I dont understand the problem

Roman55 [17]

Answer:

equation of a line:

y = mx+c

1) find the gradient, m

$m = \frac{y2 - y1}{x2 - x1}$

$m = \frac{ - 4 - ( - 4)}{ - 10 - 1}$

$m = 0$

2) find y-intercept, c using coordinate (1,-4)

y = mx + c

-4 = 0(1) + c

c = -4

the equation of line:

y = mx+c

y = 0(x) + c

y = c

y = -4

6 0

3 years ago

You deposit $5000 in an account earning 8% interest compounded monthly. How much will you have in the account in 15 years?

Rasek [7]

$\bf ~~~~~~ \textit{Compound Interest Earned Amount} \\\\ A=P\left(1+\frac{r}{n}\right)^{nt} \quad \begin{cases} A=\textit{accumulated amount}\\ P=\textit{original amount deposited}\dotfill &\$5000\\ r=rate\to 8\%\to \frac{8}{100}\dotfill &0.08\\ n= \begin{array}{llll} \textit{times it compounds per year}\\ \textit{monthly, thus twelve} \end{array}\dotfill &12\\ t=years\dotfill &15 \end{cases}$