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3 years ago

Please help!! Find the zeros and vertex of the parabola

Mathematics

1 answer:

Sergeeva-Olga [200]3 years ago

4 0

Answer:

Vertex: (1/2, 16)

Zeros: t = − 1 /2, 3/2

Step-by-step explanation:

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What are the slope and y intercept of the line described by the equation y=-4/5x -10

rodikova [14]

The slope is $-\frac{4}{5}$ and y-intercept is -10

Step-by-step explanation:

Given equation is:

$y = -\frac{4}{5}x-10$

The given equation is in slope-intercept form

General form of slope-intercept form of equation of line is:

$y = mx+b$

Here

m is the slope

b is the y-intercept

So,

comparing both equation we get

The coefficients of x

$m = -\frac{4}{5}$

And

$b = -10$

Hence,

The slope is $-\frac{4}{5}$ and y-intercept is -10

Keywords: Slope, equation of line

Learn more about equation of line at:

#LearnwithBrainly

7 0

3 years ago

A box has dimensions of 14 inches long 1.5 feet wide and 7 inches high what is the volume of the box the formula for the volume

kirza4 [7]

1.5ft to inches = 0.125 inches
14×0.125×7= 12.25 inches cubed

8 0

3 years ago

Read 2 more answers

Sandy lives 200 m from school. One

lisov135 [29]

Answer:

Following are the solution to the given question:

Step-by-step explanation:

For question 1:

Calculating the speed between A and B:

$Time\ (t)= 1 \ hours\\\\Distance\ (d)= 10 \ km\\\\V=\frac{d}{t}=\frac{10}{1}= 10 \ kmph$

For question 2:

Calculating the speed between B and C:

$Time\ (t)= (2-1)=1 \ hours\\\\Distance\ (d)= (15-10)= 5\ km\\\\V=\frac{d}{t}=\frac{5}{1}= 5 \ kmph\\$

For question 3:

In this the speed between A and B is the double to speed between B and C.

3 0

3 years ago

Mrs. 1 ran 10 meters in 3.8 seconds. Mrs. 2 ran 10 meters in 3.2 seconds.<br> Who ran faster?

Margarita [4]

Answer:

Mrs. 2

Step-by-step explanation:

3.2 seconds is shorter than 3.8 seconds

8 0

3 years ago

Read 2 more answers

A random sample of 36 students at a community college showed an average age of 25 years. Assume the ages of all students at the

Pavel [41]

Answer:

98% confidence interval for the average age of all students is [24.302 , 25.698]

Step-by-step explanation:

We are given that a random sample of 36 students at a community college showed an average age of 25 years.

Also, assuming that the ages of all students at the college are normally distributed with a standard deviation of 1.8 years.

So, the pivotal quantity for 98% confidence interval for the average age is given by;

P.Q. = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample average age = 25 years

$\sigma$ = population standard deviation = 1.8 years

n = sample of students = 36

$\mu$ = population average age

So, 98% confidence interval for the average age, $\mu$ is ;

P(-2.3263 < N(0,1) < 2.3263) = 0.98

P(-2.3263 < $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < 2.3263) = 0.98

P( $-2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < ${\bar X - \mu}$ < $2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

P( $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ) = 0.98

98% confidence interval for $\mu$ = [ $\bar X - 2.3263 \times {\frac{\sigma}{\sqrt{n} }$ , $\bar X +2.3263 \times {\frac{\sigma}{\sqrt{n} }$ ]

= [ $25 - 2.3263 \times {\frac{1.8}{\sqrt{36} }$ , $25 + 2.3263 \times {\frac{1.8}{\sqrt{36} }$ ]

= [24.302 , 25.698]

Therefore, 98% confidence interval for the average age of all students at this college is [24.302 , 25.698].

8 0

3 years ago