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3 years ago

On a final exam, each multiple-choice question is worth 4 points and each word problem is worth 8 points. Lorenzo needs

Mathematics

2 answers:

Elena-2011 [213]3 years ago

7 0

Answer:

4x + 8y = 50 or 4x + 8y > 50

Step-by-step explanation:

To find which equation show the exact answer, Becasue he need at least 50 poitns to have a B, so:

4x + 8y must larger or equal to 50

In these choice, some I can't figure out what it means but the correct answer for this is:

4x + 8y = 50 or 4x + 8y > 50

Hope this help you :3

mr_godi [17]3 years ago

6 0

Answer:

4x + 8y ≥ 50

Step-by-step explanation:

Since x represents the multiple choice questions, and each of them is worth 4 points, the total points because of this type of questions is

and y represents the word problems, and each of this is worth 8 point, so the total points because of this type of questions is

Lorenzo needs at leat 50 points: this means 50 points or more, so the sum of the points he gets for each type of question must be equal to 50 points or more.

This is represented by the inequation:

4x + 8y ≥ 50

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If we wanted to create a new 90% confidence interval from a different sample for the proportion of those with a two on one date

slava [35]

Answer: 271

Step-by-step explanation:

The formula we use to find the sample size is given by :-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

, where $z_{\alpha/2}$ is the two-tailed z-value for significance level of $(\alpha)$

p = prior estimation of the proportion

E = Margin of error.

If prior estimation of the proportion is unknown, then we take p= 0.5 , the formula becomes

$n=0.5(1-0.5)(\dfrac{z_{\alpha/2}}{E})^2$

$n=0.25(\dfrac{z_{\alpha/2}}{E})^2$

Given : Margin of error : E= 0.05

Confidence level = 90%

Significance level $\alpha=1-0.90=0.10$

Using z-value table , Two-tailed z-value for significance level of $0.10$

$z_{\alpha/2}=1.645$

Then, the required sample size would be :

$n=0.25(\dfrac{1.645}{0.05})^2$

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$n=270.6025\approx271$

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3 years ago