Round this to the nearest hundredth place please!!!

which experimental probability from coach nelson's experiment equals the theorectical probability? what is this probability?

Lilit [14]

Answer:

QUESTION:

which experimental probability from coach nelson's experiment equals the theorectical probability? what is this probability?

ANSWER:

This will explain everything to you.

https://www.khanacademy.org/math/cc-seventh-grade-math/cc-7th-probability-statistics/cc-7th-theoretical-and-experimental-probability/v/comparing-theoretical-to-experimental-probabilites

Step-by-step explanation:

Hope that this helps you out! :)

If you have any questions please put them in the comment section below this answer.

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7 0

3 years ago

1. The structure of the supports for the hill consists of steel beams that

Darina [25.2K]

a) Parallel for both line segments have the same slope = -5/4

b) 55º

Step-by-step explanation:

a) To verify the parallelism of these line segments, It's necessary to find the slope of each one. Parallel lines share the same slope value:

$m_{BC}=\frac{0-150}{120-0}=-\frac{-5}{4}\\m_{AD}=\frac{0-125}{100-0}=-\frac{5}{4}$

b) If ∠DAB = 125°, what is the measure of ∠CBA? Justify your reasoning.

Those beams do not form a parallelogram because:

Using the formula of distance between those points

$d=\sqrt{(y_{2}-y_1)^{2}-(x_{2}-x_{1})^2$

$\overline {AB}=25\: \: \: \overline{CD}=20\: \: \overline{AD}=160 \: \: \overline{BC}=192$

So this is a trapezoid. And two adjacent angles within a trapezoid are supplementary, so If ∠DAB = 125º then 180 - ∠CBA=55º

3 0

3 years ago

Find the angle which 7x+5y=35 makes with y axis

7nadin3 [17]

answer is

$x = \frac{ - 5}{7}y + 5$

8 0

3 years ago

The half-life of radioactive strontium-90 is approximately 28 years. In 1960, radioactive strontium-90 was released into the atm

pentagon [3]

Answer: It will take about 77 years (76.8 years).

Step-by-step explanation: Time is given by the equation: N=No.e^kt. You know that the concentration has to be 15% from the original so N=15 and No=100. To determine k, you have to look at the half-life. k=ln(0.5)/half-life time. For 28 years, k=-0.0247. Replacing it back into the N=No.e^kt equation, the answer will be 76.8 years.

6 0

3 years ago

A certain test preparation course is designed to help students improve their scores on the GMAT exam. A mock exam is given at th

Bogdan [553]

Answer:

95% confidence interval for the average net change is between a lower limit of 11.709 and an upper limit of 30.921.

The critical value that should be used in constructing the confidence interval is 4.303.

Step-by-step explanation:

Confidence interval is given as mean +/- margin of error (E)

mean = (17+20+26)/3 = 63/3 = 21

sd = sqrt[((17-21)^2 + (20-21)^2 + (26-21)^2) ÷ 3] = sqrt(42÷3) = sqrt(14) = 3.74

n = 3

degree of freedom = n-1 = 3-1 = 2

confidence level (C) = 95% = 0.95

significance level = 1 - C = 1 - 0.95 = 0.05 = 5%

critical value (t) corresponding to 2 degrees of freedom and 5% significance level is 4.303.

The critical value is 4.303

E = t×sd/√n = 4.303×3.74/√3 = 9.291

Lower limit of mean = mean - E = 21 - 9.291 = 11.709

Upper limit of mean = mean + E = 21 + 9.291 = 30.291

95% confidence interval is (11.709, 30.391)

8 0

3 years ago