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3 years ago

I neeeddddd helpppppppp plllllzzzz

Mathematics

1 answer:

vesna_86 [32]3 years ago

6 0

Step-by-step explanation:

-3x + 7y = 5x + 2y (you subtract the 2y from both sides)

-3× + 5y = 5x (then add 3x to both sides)

5y = 8x (use x = - 5)

5y = 8x - 5 = -40 (both sides are divided by 5)

y = -8

answer is (-5, -8)

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Find the midpoint of the line segment with the endpoints (1, 0) and (2, -10).

Mariulka [41]

Answer:

$\bigl(\frac{3}{2},-5\bigr)$

or in decimal form

$(1.5,-5)$

Step-by-step explanation:

The midpoint of the line with endpoints $(x_1,y_1)$ and $(x_2,y_2)$ is $\bigl(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigr)$ . just take the average between the points

so given the points (1,0) and (2,-10)

$x_1=1$

$y_1=0$

$x_2=2$

$y_2=-10$

the midpoint is found as follows:

$\bigl(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigr)$

$\bigl(\frac{1+2}{2},\frac{0-10}{2}\bigr)$

$\bigl(\frac{3}{2},\frac{-10}{2}\bigr)$

$\bigl(\frac{3}{2},-5\bigr)$

or in decimal form

$(1.5,-5)$

7 0

3 years ago

A shoe manufacturer collected data regarding men's shoe sizes and found that the distribution of sizes exactly fits the normal c

Lina20 [59]

Answer: 97.72%

Step-by-step explanation:

Given : A shoe manufacturer collected data regarding men's shoe sizes and found that the distribution of sizes exactly fits the normal curve.

Let x be the random variable that represents the shoe sizees.

Also, The population mean = $\mu=11$ ; Standard deviation: $\sigma=1.5$

Formula for z:-

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

Put x= 8, we get

$z=\dfrac{8-11}{1.5}=-2$

Now, the probability that the male shoe sizes are greater than 8 :-

$P(x>8)=P(z>-2)=1-P(z\leq-2)\\\\=1-0.0227501=0.9772499\approx0.9772$

Hence, the percent of male shoe sizes are greater than 8 is 97.72%.

7 0

3 years ago

According to a Los Angeles Times study of more than 1 million medical dispatches from to , the response time for medical aid var

-BARSIC- [3]

Answer:

A)Mean :10.65

Median =10.7

Mode : 10.7

B)Range = 3.5

Standard deviation :0.89916

C)The response time of 8.3 minutes should be considered an outlier in comparison to the other response times

Step-by-step explanation:

Data : 11.8 ,10.3, 10.7, 10.6, 11.5, 8.3, 10.5, 10.9, 10.7, 11.2

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\Mean = \frac{11.8 +10.3+ 10.7+ 10.6+ 11.5+ 8.3+ 10.5+ 10.9+ 10.7+ 11.2}{10}$

Mean = 10.65

Median: The mid value of the data

Data in ascending order

8.3

10.3

10.5

10.6

10.7

10.9

11.2

11.5

11.8

n=10(even)

$Median = \frac{(\frac{n}{2})\text{th term}+(\frac{n}{2}+1)\text{th term}}{2}\\Median = \frac{(\frac{10}{2})\text{th term}+(\frac{10}{2}+1)\text{th term}}{2}\\Median = \frac{10.7+10.7}{2}\\Median = 10.7$

Mode : the most occurring frequency

10.7 is occurring twice while others are occurring once

So, Mode is 10.7

B) Range = Maximum - Minimum=11.8-8.3=3.5

Standard deviation : $\sqrt{\frac{\sum(x-\bar{x})^2}{n}}$

Standard deviation : $\sqrt{\frac{(11.8-10.65)^2+(10.3-10.65)^2+......+(10.9-10.65)^2+(10.7-10.65)^2+(11.2-10.65)^2}{10}}$

Standard deviation :0.89916

8.3

,10.3

,10.5

,10.6

,10.7

,10.9

,11.2

,11.5

,11.8

For Q1 ( Median of lower quartile )

8.3

,10.3

,10.5

,10.6

,10.7

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=10.5$

For Q3( Median of Upper quartile )

10.7

,10.9

,11.2

,11.5

,11.8

$Median = \frac{n+1}{2}\text{th term} =\frac{5+1}{2}=3 \text{rd term}=11.2$

IQR = Q3-Q1=11.2-10.5=0.7

Range :(Q1-1.5IQR, Q3-1.5IQR)

Range : $(10.5-1.5 \times 0.7, 11.2-1.5 \times 0.7)$

Range :(9.45, 10.15)

8.3 does not lie in this interval

So, the response time of 8.3 minutes should be considered an outlier in comparison to the other response times

3 0

3 years ago

The stock of Company A lost $3.63 throughout the day and ended at a value of

dusya [7]

Answer:

6% decline

Step-by-step explanation:

Original stock price = $3.63 + $56.87 = $60.50

Percentage decline = 60.50 - 56.87 / 60.50

= 3.63 / 60.50

= 0.06

= 6% decline

6 0

2 years ago

Three siblings are participating in a family-friendly running event. - The oldest sibling begins at the start line of the race a

Aleonysh [2.5K]

Answer:

What is le question

Step-by-step explanation:

3 0

3 years ago