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2 years ago

The solution to the given system of linear equations lies in which quadrant?

Mathematics

2 answers:

goblinko [34]2 years ago

8 0

Answer:

The solution lies in quadrant IV.

Step-by-step explanation:

Rina8888 [55]2 years ago

5 0

X-3y=6
x+y=2
this is an substitution problem
so first you can do is rewrite the problem by subjection one variable
x=3y+6
then substitute this in the other proble
x+y=2
(3y+6)+y=2
4y+6=2
4y=2-6
4y=-4
y=-1
then substitute the no. in the original equation.
x=3y+6
x=3(-1)+6
x=-3+6
x=3
now you got the intercepts and you draw the line and check.
it's in the IV quadrant

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Elliot has a total of 26 books. He has 12 more fiction books than nonfiction books. Let x represent the number of fiction books

Marta_Voda [28]

Given:

Elliot has a total of 26 books. He has 12 more fiction books than nonfiction books.

To Find:

A system of linear equations represents the situation.

Answer:

$x+y=26\\\\x=y+12$

Step-by-step explanation:

We are given that x represents the number of fiction books and y is the number of non-fiction books.

We are also given that the total number of books Elliot has is 26 which includes both fiction and non-fiction. So, we may write

$x+y=26$

Next, we are given that there are 12 more fiction books than non-fiction books. This means, the fiction books are more in number and so, we may write

$x=y+12$

So, the total system of equations can be represented as

$x+y=26\\\\x=y+12$

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2 years ago

Read 2 more answers

Amy is just learning how to rock climb. Her instructor takes her to a 26 ft climbing wall for her first time. She climbs up 5 ft

boyakko [2]

Let's compute the speeds as she goes up and down of the climbing wall. Speed is the ratio of distance to time.

Speed going up = 5 ft/(2min * 60 s/1 min) = 1/24 ft/s
Speed going down = 2 ft/10 s = 0.2 ft/s
Net speed = 1/24 ft/s - 0.2 ft/s = 5/24 ft/s

Using this net speed, we can already calculate for the total time:

Speed = Distance/Time
5/24 ft/s = 26 ft/Time
Time = 124.8 seconds or 2.08 minutes

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2 years ago

Round to the nearest hundred thousand <br> 89, 659

victus00 [196]

Answer:

The nearest hundred thousand of the provided number is 100,000.

Step-by-step explanation:

Consider the provided number 89, 659

According to place value.

Hundred Th. Ten Th. Thousands hundreds Tens Ones

100,000 10,000 1000 100 10 1

We need to round it to the nearest hundred thousand.

The provided number can be written as 089,659

The digit at the hundred thousand place is 0.

The rule of rounding a number is:

If 0, 1, 2, 3, or 4 follow the number, then no need to change the rounding digit.

If 5, 6, 7, 8, or 9 follow the number, then rounding digit rounds up by one number.

Here, the number at the ten thousands place is 8, so to round up the number increase the digit of hundred thousand place by 1.

The digit at the hundred thousand place is 0 so increase it by 1.

Thus, the number can be rounded to the nearest hundred thousand is shown as:

100,000

The nearest hundred thousand of the provided number is 100,000.

8 0

1 year ago

Read 2 more answers

A car insurance company suspects that the younger the driver is, the more reckless a driver he/she is. They take a survey and gr

erastovalidia [21]

Answer:

The confidence interval for the difference in proportions is

$-0.028\leq p_1-p_2 \leq 0.096$

No. As the 95% CI include both negative and positive values, no proportion is significantly different from the other to conclude there is a difference between them.

Step-by-step explanation:

We have to construct a confidence interval for the difference of proportions.

The difference in the sample proportions is:

$p_1-p_2=x_1/n_1-x_2/n_2=(183/217)-(322/398)=0.843-0.809\\\\p_1-p_2=0.034$

The estimated standard error is:

$\sigma_{p_1-p_2}=\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2} } \\\\\sigma_{p_1-p_2}=\sqrt{\frac{0.843*0.157}{217}+\frac{0.809*0.191}{398} } \\\\\sigma_{p_1-p_2}=\sqrt{0.000609912+0.000388239}=\sqrt{0.000998151} \\\\ \sigma_{p_1-p_2}=0.0316$

The z-value for a 95% confidence interval is z=1.96.

Then, the lower and upper bounds are:

$LL=(p_1-p_2)-z*\sigma_p=0.034-1.96*0.0316=0.034-0.062=-0.028\\\\\\UL=(p_1-p_2)+z*\sigma_p=0.034+1.96*0.0316=0.034+0.062=0.096$

The confidence interval for the difference in proportions is

$-0.028\leq p_1-p_2 \leq 0.096$

<em>Can it be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group?</em>

No. It can not be concluded that there is a difference in the proportion of drivers who wear a seat belt at all times based on age group, as the confidence interval include both positive and negative values.

This means that we are not confident that the actual difference of proportions is positive or negative. No proportion is significantly different from the other to conclude there is a difference.

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2 years ago

Monica has a bag of marbles. There are 4 red marbles and 7 blue marbles and 5 yellow marbles in a bag. Monica will randomly pick

natima [27]

Answer:

$\frac{7}{64}\approx 0.11$

Step-by-step explanation:

We have been given that there are 4 red marbles and 7 blue marbles and 5 yellow marbles in a bag. Monica will randomly pick two marbles out of the bag replacing the first marble before picking the second marble.

Since Monica will replace the first marble before picking the second marble, therefore, probability of both events will be independent and probability of occurring one event will not affect the probability of second event's occurring.

Since the probability of two independent compound events is always the product of probabilities of both events.

$P(\text{A and B})=P(A)*P(B)$