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3 years ago

6

Tina is saving to buy a notebook computer. She has two options. The first option is to put $400 away initially and save $20 ever

y month. The second option is to put $300 away initially and save $40 every month. After how many months would Tina save the same amount using either option? How much would she save with either option?
After [ ] months, Tina will save $[ ] with either option.

1 answer:

Mrrafil [7]3 years ago

7 0

Her savings amounts after x months will be ...

... 400 + 20x

or

... 300 + 40x

You want to find x such that these are the same amount. So, we equate these values and solve for x.

... 400 + 20x = 300 + 40x

... 0 = -100 + 20x . . . . . . . . . subtract the left side

... 0 = -5 + x . . . . . . . . . . . . . divide by the coefficient of x

... 5 = x . . . . . . . . . . . . . . . . add the opposite of the constant

Using the first option, we find the savings amount to be

... 400 + 20·5 = 500

After 5 months, Tina will save $500 with either option.

_____

Once you have the expressions for the options, it is convenient to let a graphing calculator tell you the point where they are the same.

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Jane and Nancy were both awarded National Merit Scholarships (for academic excellence), and both love mathematics. Jane breezed

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Answer:

c. Inductive and Strong

Step-by-step explanation:

In inductive reasoning, provided data is analyzed in order to reach a conclusion. In this case, the argument provides data regarding Jane and Nancy's awards and their love for mathematics and then draws a conclusion regarding Nancy's performance in a particular class, this is an example of inductive reasoning.

As for the strength of the argument, it is plausible to infer that Jane and Nancy have similar mathematics skills since they both love calculus and excel academically. Therefore, if Jane does well in the calculus class, it is a strong argument to say that Nancy does as well.

The answer is :

c. Inductive and Strong

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3 years ago

A statistics student wants to compare his final exam score to his friend's final exam score from last year; however, the two exa

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Answer:

$z= \frac{85-70}{10}=1.5$

$z= \frac{45-35}{5}=2$

So then the correct answer for this case is:

B) Our student, Z= 1.50; his friend, Z=2.00.

Step-by-step explanation:

Assuming this complete question:

A statistics student wants to compare his final exam score to his friend's final exam score from last year; however, the two exams were scored on different scales. Remembering what he learned about the advantages of Z scores, he asks his friend for the mean and standard deviation of her class on the exam, as well as her final exam score. Here is the information:

Our student: Final exam score = 85; Class: M = 70; SD = 10.

His friend: Final exam score = 45; Class: M = 35; SD = 5.

The Z score for the student and his friend are:

A) Our student, Z= -1.07; his friend, Z= -1.14.

B) Our student, Z= 1.50; his friend, Z=2.00.

C) Our student, Z= 1.07; his friend, Z= -1.14.

D) Our student, Z= 1.07; his friend, Z= 1.50

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution

Let X the random variable that represent the scores for our student, and for this case we know that:

$E(X)= \mu =70, SD_X=\sigma=10$

The z score is given by:

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

If we use this we got:

$z= \frac{85-70}{10}=1.5$

Let Y the random variable that represent the scores for his friend, and for this case we know that:

$E(Y)= \mu =35, SD_Y=\sigma=5$

The z score is given by:

$z=\frac{y-\mu}{\sigma}$

If we use this we got:

$z= \frac{45-35}{5}=2$

So then the correct answer for this case is:

B) Our student, Z= 1.50; his friend, Z=2.00.

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3 years ago

How many Solutions does this system have? (1 point)

mixas84 [53]

The given system of equation that is $2x+y=3$ and $6x=9-3y$ has infinite number of solutions.

Option -C.

<u>Solution:</u>

Need to determine number of solution given system of equation has.

$\begin{array}{l}{2 x+y=3} \\\\ {6 x=9-3 y}\end{array}$

Let us first bring the equation in standard form for comparison

$\begin{array}{l}{2 x+y-3=0} \\\\ {6 x+3 y-9=0}\end{array}$

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

To check how many solutions are there for system of equations $a_{1} x+b_{1} y+c_{1}=0 \text{ and }a_{2} x+b_{2} y+c_{2}=0$ , we need to compare ratios of $\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}$

In our case,

$a_{1} = 2, b_{1}= 1\text{ and }c_{1}= -3$

$a_{2} = 6, b_{2} = 3,\text{ and }c_{2} = -9$

$\begin{array}{l}{\Rightarrow \frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3}} \\\\ {\Rightarrow \frac{b_{1}}{b_{2}}=\frac{1}{3}} \\\\ {\Rightarrow \frac{c_{1}}{c_{2}}=\frac{-3}{-9}=\frac{1}{3}} \\\\ {\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{3}}\end{array}$

As $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ , so given system of equations have infinite number of solutions.

Hence, we can conclude that system has infinite number of solutions.

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3 years ago

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The answer should be 26 hours

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3 years ago

A bag contains 5 red marbles, 4 blue marbles and 6 green marbles. If three marbles

Nina [5.8K]

Answer:

The probability of drawing three red marbles is 1/27, or about 3.71%

Step-by-step explanation:

Add all the marbles:

5 + 4 + 6 = 15

Find favourable outcomes:

5

The probability for drawing a red marble is:

5/15 or 1/3

Multiply this ×3 to find the probability of drawing three red marbles.

1/3 × 1/3 × 1/3 = 1/27

1 ÷ 27 = 0.0370370370 ... repeating

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3 years ago