Pls help ASAP!!!!!!

Jonathons piggy bank contains 20 nickels 30 quarters and 50 one dollar coins. He picks 20 coins from the bank at random

fredd [130]

Answer:

All in all, Jonathan's piggy bank contains 100 coins. Among these coins, only 50 are one-dollar coins. Therefore, the theoretical probability of picking one-dollar coin from the piggy bank is equal to 50/100 or 1/2.

Similarly, from the experiment, 20 coins were picked and among these there are 12 one-dollar coins. The answer to the second question is therefore 12/20 or 3/5.

Step-by-step explanation:

4 0

1 year ago

If the sample space S is an infinite set, does this necessarily imply that any random variable X defined from S will have an inf

alekssr [168]

Answer: no

Step-by-step explanation: see attachment

4 0

1 year ago

The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillat

Greeley [361]

Answer:

1) $L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

2) $T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

3) $T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

Step-by-step explanation:

Part 1

For this case we know the following info: The length, l cm, of a simple pendulum is directly proportional to the square of its period (time taken to complete one oscillation), T seconds.

$L \propto T^2$

Using the condition given:

$2.205 m = K (3)^2$

$K = 0.245 \approx \frac{g}{4\pi^2}$

So then if we want to create an equation we need to do this:

$L = K T^2$

With K a constant. For this case the period of a pendulumn is given by this general expression:

$T = 2\pi \sqrt{\frac{L}{g}}$

Where L is the length in m and g the gravity $g = 9.8 \frac{m}{s^2}$ .

Part 2

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

If we square both sides of the equation we got:

$T^2 = 4 \pi^2 \frac{L}{g}$

And solving for L we got:

$L = \frac{g T^2}{4 \pi^2}$

Replacing we got:

$L =\frac{9.8 \frac{m}{s^2} (5s)^2}{4 \pi^2} = 6.206m$

Part 3

For this case using the function in part a we got:

$T = 2\pi \sqrt{\frac{L}{g}}$

Replacing we got:

$T = 2\pi \sqrt{\frac{0.98m}{9.8\frac{m}{s^2}}}= 1.987 s$

8 0

2 years ago

The paired data below consist of the temperatures on randomly chosen days and the amount a certain kind of plant grew (in millim

rjkz [21]

Answer:

Step-by-step explanation:

Given is a paired data which consist of temperatures (X in mm) and growth

We have to find the linear correlation i.e. the measure of association between these two variables.

x y xy x^2 y^2

62 36 2232 3844 1296

76 39 2964 5776 1521

50 50 2500 2500 2500

51 13 663 2601 169

71 33 2343 5041 1089

46 33 1518 2116 1089

51 17 867 2601 289

44 6 264 1936 36

79 16 1264 6241 256

Mean 58.88888889 27 1623.888889 3628.444444 916.1111111

cov 33.88888889

std dev x 13.43916333 14.50861813

sx *sy

r 0.195529176

Hence we find that correlation coefficient 0.1955.

7 0

2 years ago

What is the following quotient 3 sqrt 8/ 4 sqrt 6

adell [148]

We have that
(3√8)/(4√6)

we know that
√8---------> √(2³)-----> 2√2
so
(3√8)/(4√6)=(3*[2√2])/(4√6)---> 6√2/(4√6)
√6=√(2*3)---> √2*√3
6√2/(4√6)=6√2/(4√2*√3)----> 6/(4√3)----> 6/(4√3)*(√3/√3)-----> 6√3/(4*3)----> √3/2

the answer is
√3/2

7 0

1 year ago

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