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2 years ago

Is the relationship between the 7s in 7742 and the 7s in 7785 different in any way

Mathematics

2 answers:

marysya [2.9K]2 years ago

7 0

No it is not because the sevens are in the same place in both numbers

likoan [24]2 years ago

7 0

No, ther is no difference (random) but if you round 7,785 it would be greater than 7,742
no, both the 77 is in the same place value

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Cora is playing a game that involves flipping three coins at once. Let the random variable HHH be the number of coins that land

arlik [135]

Answer:

0.875

Step-by-step explanation:

P(H=0) = 0.125

P(H=1) = 0.375

P(H=2) = 0.375

P(H=3) = 0.125

P(H<3) = P(H=0) + P(H=1) + P(H=2)

P(H<3) = 0.125 + 0.375 + 0.375

P(H<3) = 0.875

4 0

2 years ago

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Use the normal approximation to the binomial distribution to answer this question. Fifteen percent of all students at a large un

Nataly_w [17]

Answer: 0.1289

Step-by-step explanation:

Given : The proportion of all students at a large university are absent on Mondays. : $p=0.15$

Sample size : $n=12$

Mean : $\mu=np=12\times0.15=1.8$

Standard deviation = $\sigma=\sqrt{np(1-p)}$

$\Rightarrow\ \sigma=\sqrt{12(0.15)(1-0.15)}=1.23693168769\approx1.2369$

Let x be a binomial variable.

Using the standard normal distribution table ,

$P(x=4)=P(x\leq4)-P(x\leq3)$ (1)

Z score fro normal distribution:-

$z=\dfrac{x-\mu}{\sigma}$

For x=4

$z=\dfrac{4-1.8}{1.2369}\approx1.78$

For x=3

$z=\dfrac{3-1.8}{1.2369}\approx0.97$

Then , from (1)

$P(x=4)=P(z\leq1.78)-P(z\leq0.97)\\\\=0.962462-0.8339768\approx0.1289$

Hence, the probability that four students are absent = $0.1289$

3 0

2 years ago

The parent cosecant function is shifted 4 units right and 3 units up. Which of the following is the graph of the transformed fun

snow_lady [41]

We have that

using a graph tool--------------> graph the <span> cosecant function
</span>see the attached figure

the answer is the option B

4 0

2 years ago

Read 2 more answers

The table shows the number of grapes eaten over several minutes. What is the rate of change for the function on the table? 15 gr

Mkey [24]

Answer:

1. 15 grapes eaten per minute.

Step-by-step explanation:

We have been given a table that shows the number of grapes eaten over several minutes. We are asked to find the rate of change of our given function.

Since we know that rate of change of a function can be found by vertical change of the function divided by horizontal change of the function.

$\text{Rate of change of a function}=\frac{\text{Vertical change}}{\text{Horizontal change}}$

$\text{Rate of change of a function}=\frac{y_2-y_1}{x_2-x_1}$

$y_2-y_1$ =Difference between two y-coordinates.

$x_2-x_1$ =Difference between corresponding x-coordinates of y-coordinates.

Upon substituting coordinates of any two points in above formula we will get,

$\text{Rate of change of the function}=\frac{60-15}{4-1}$

$\text{Rate of change of the function}=\frac{45}{3}$

$\text{Rate of change of the function}=15$

We can see that rate of change of our given function is 15 grapes eaten per minute and 1st option is the correct choice.

7 0

2 years ago

Read 2 more answers

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