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netineya [11]
1 year ago
12

smartbook 2.0 allows you to start your active learning experience by either reading or by jumping right in to start answering qu

estions.
SAT
1 answer:
romanna [79]1 year ago
3 0

Yes it is right that smartbook 2.0 allows you to start your active learning experience by either reading or by jumping right in to start answering questions.

Given that smartbook 2.0 allows you to start your active learning experience by either reading or by jumping right in to start answering questions.

We are required to tell the whether the statement about smartbook 2.0 is right or wrong.

SmartBook 2.0 basically creates a baseline of student knowledge and focuses their time on knowledge gaps.SmartBook 2.0 also helps students better understand what they know and what they don't know.

So,it is right in saying that smartbook 2.0 allows you to start your active learning experience by either reading or by jumping right in to start answering questions.

Hence it is right in saying that smartbook 2.0 allows you to start your active learning experience by either reading or by jumping right in to start answering questions.

Learn more about smartbook 2.0 at brainly.com/question/14436827

#SPJ4

You might be interested in
Antique Accents tracks their daily profits and has found that the distribution of profits is approximately normal with a mean of
mezya [45]

Answer;

a) 0.434

b) 0.983

c) 0.367

Explanation:

The exact question with the given parameters wasn't obtained online, but the same question, albeit with different parameters is then obtained. Hopefully, this Helps to solve the complete question with the required parameters.

Antique Accents tracks their daily profits and has found that the distribution of profis is approximately normal with a mean of $17,700.00 and a standard deviation of about $900.00. Using this information, answer the following questions For full marks your answer should be accurate to at least three decimal places. Compute the probability that tomorrow's profit will be

a) less than $16,791 or greater than $18,231

b) greater than $15,783

c) between $17,997 and $20,130

Solution

This is a normal distribution problem with

Mean = μ = $17,700

Standard deviation = σ = $900

a) less than $16,791 or greater than $18,231. P(x < 16,791) or P(X > 18,231) = P(X < 16,791) + P(x > 18,231)

We first standardize 16,791 and 18,231

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 16791

z = (x - μ)/σ = (16791 - 17700)/900 = - 1.01

For 18231

z = (x - μ)/σ = (18231 - 17700)/900 = 0.59

To determine the required probability

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59)

We'll use data from the normal probability table for these probabilities

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59)

P(z < -1.01) = 0.15625

P(z > 0.59) = 1 - (z ≤ 0.59) = 1 - 0.7224 = 0.2776

P(X < 16,791) + P(x > 18,231) = P(z < -1.01) + P(z > 0.59) = 0.15625 + 0.2776 = 0.43385 = 0.434 to 3 d.p

b) greater than $15,783. P(x > 15783)

We standardize 15783

z = (x - μ)/σ = (15783 - 17700)/900 = -2.13

To determine the required probability

P(x > 15783) = P(z > -2.13)

We'll use data from the normal probability table for this probability

P(x > 15783) = P(z > -2.13) = 1 - P(z ≤ - 2.13)

= 1 - 0.01659 = 0.98341 = 0.983 to 3 d.p.

c) between $17,997 and $20,130.

P(17,997 < x < 20,130)

We first standardize 17,997 and 20,130

The standardized score for any value is the value minus the mean then divided by the standard deviation.

For 17,997

z = (x - μ)/σ = (17,997 - 17700)/900 = 0.33

For 20,130

z = (x - μ)/σ = (20,130 - 17700)/900 = 2.70

To determine the required probability

P(17,997 < x < 20,130) = P(0.33 < x < 2.70)

We'll use data from the normal probability table for these probabilities

P(17,997 < x < 20,130) = P(0.33 < x < 2.70)

= P(z < 2.70) - P(z < 0.33)

= 0.99653 - 0.62930

= 0.36723 = 0.367 to 3 d.p.

Hope this Helps!!!

3 0
3 years ago
Find the residual values, and use the graphing calculator tool to make a residual plot.
finlep [7]

Answer:

  • No, the points are evenly distributed about the x-axis.

Explanation:

<u>1. Write the table with the data:</u>

  x      given     predicted       residual

  1       - 3.5         - 1.1

  2      - 2.9           2

  3       - 1.1            5.1

  4        2.2           8.2

  5        3.4           1.3

<u>2. Complete the column of residuals</u>

The residual is the observed (given) value - the predicted value.

  • residual = given - predicted.

Thus, the complete table, with the residual values are:

  x      given     predicted       residual

  1       - 3.5         - 1.1                 - 2.4

  2      - 2.9           2                  - 4.9

  3       - 1.1            5.1                - 6.2

  4        2.2           8.2               - 6.0

  5        3.4           1.3                   2.1

<u>3. Residual plot</u>

You must plot the last column:

  x      residual

  1       - 2.4

  2      - 4.9

  3       - 6.2

  4       - 6.0

  5        2.1

See the plot attached.

<em>Does the residual plot show that the line of best fit is appropriate for the data?</em>

Ideally, a residual plot for a line of best fit that is appropiate for the data must not show any pattern; the points should be randomly distributed about the x-axis.

But the points of the plot are not randomly distributed about the x-axis:  there are 4 points below the x-axis and 1 point over the x-axis: there are more negative residuals than positive residuals.  This is a pattern. Also, you could say that they show a curve pattern, which drives to the same conclusion: the residual plot shows that the line of best fit is not appropiate for the data.

Thus, the conclusion should be: No, the points have a pattern.

  • 1. "<em>Yes, the points have no pattern</em>": false, because as shown, the points do have a pattern, which makes the residual plots does not show that the line of best fit is appropiate for the data.

  • 2. "<em>No, the points are evenly distributed about the x-axis</em>": true. As already said the points have a pattern. It is a curved pattern, and this <em>shows the line of best fit is not appropiate for the data.</em>

  • 3. "<em>No, the points are in a linear pattern</em>": false. The points are not in a linear pattern.

  • 4. "<em>Yes, the points are in a curved pattern</em>": false. Because the points are in a curved pattern, the residual plot shows that the line of best fit is not appropiate for the data.

3 0
4 years ago
Read 2 more answers
What best explains the label that the student should use on the y-axis? reaction rate, as the rate of forward and backward react
Nata [24]

Answer:

5+5=10

Explanation:

Now to get that answer you need two hands make sure they both have five fingers now put your hands toughter now then count the fingers and if you count them there should be 10 unless you ate one yesterday like I did.

8 0
2 years ago
I'll give anyone brainliest if you can tell me what the point of getting brainliest is!
Margarita [4]

Hey there!

Brainliest is given to the best answer. Collecting brainliest helps get users to the next “level” or “rank”. The more brainliest answers you have, the quicker you’ll get to the next rank. If you go to your profile, and click your rank (it’s right under your username) a list of the different ranks will come up. If you need any more info, let me know!

Hope this helps you!

God bless ❤️

xXxGolferGirlxXx

8 0
3 years ago
Read 2 more answers
Is normal VPN apps can change your location ?​
mestny [16]
Yes, they can.
that’s why are they made.
VPN means virtual private network which hides/changes your IP address and connects you to another network which changes your location and allows you to surf privately and safely on the internet
8 0
3 years ago
Read 2 more answers
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