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3 years ago

Please help ASAP! I always give brainliest!(unless both answers are perfect) and thx

Mathematics

1 answer:

vlabodo [156]3 years ago

5 0

Answer:

This is assuming that both x and y do not equal zero, if that is necessary information. 1 minute ago Solve for x in the equation x squared minus 14 x + 31 = 63.

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For what value of “a”are the graphs of 5y = -2x + 10 and 3y = aX- 15 parallel

Pie

Answer:

a = -6/5

Step-by-step explanation:

For the graphs to be parallel the graphs should have same slope(m)

So we rewrite both our equations in the slope-intercept form then compare the slope to find the value of a like this,

$y=mx+b$

This equation is the slope-intercept form we convert both our equations in this form firstly taking equation 1

$5y=-2x+10\\\\y=\frac{-2x+10}{5} \\\\y=\frac{-2}{5}x+\frac{10}{5} \\\\y=\frac{-2}{5}x+2$

so if we compare it with y = mx + b the coefficient of x is m and hence

m= -2/5 now solving for equation 2

$3y=ax-15\\\\y=\frac{ax-15}{3} \\\\y=\frac{ax}{3}-\frac{15}{3} \\\\y=\frac{a}{3}x-5\\\\$

so here if we compare it with y = mx + b the coeffienct of x is a/3 so since parallel lines have same slope by the formula:

$m_1=m_2$

we equation both the slope to each other to find the value of a like this,

$m_1=m_2\\\\\frac{-2}{5}=\frac{a}{3}\\\\-2(3)=a(5)\\\\-6=5a\\-6/5=a$

so the value of a equals

a= -6/5

8 0

2 years ago

What are the zeros of this function?

Yakvenalex [24]

Answer: 0 and 5

Step-by-step explanation: The zeros of the function are the points were the graph pass through x - axis

5 0

3 years ago

Gwen has 3 red erasers, 4 green erasers, 2 black erasers, 5 purple erasers, and 6 yellow erasers in her book bag.

son4ous [18]

I think the probability that Gwen will reach in her book bag purple eraser is 5/20 or 1/4

5 0

2 years ago

Because of staffing decisions, managers of the Gibson-Marion Hotel are interested in the variability in the number of rooms occu

olchik [2.2K]

Answer:

a) $s^2 =30^2 =900$

b) $\frac{(19)(30)^2}{30.144} \leq \sigma^2 \leq \frac{(19)(30)^2}{10.117}$

$567.28 \leq \sigma^2 \leq 1690.224$

c) $23.818 \leq \sigma \leq 41.112$

Step-by-step explanation:

Assuming the following question: Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in the variability in the number of rooms occupied per day during a particular season of the year. A sample of 20 days of operation shows a sample mean of 290 rooms occupied per day and a sample standard deviation of 30 rooms

Part a

For this case the best point of estimate for the population variance would be:

$s^2 =30^2 =900$

Part b

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$