Find zeros of quadratic function

Graph the following system of linear inequalities. Identify at least two points in the solution: y < 5 - 2x | x + 5y > -7

romanna [79]

Answer:

(1,2) and (2,-1)

Step-by-step explanation:

we have

$y< 5-2x$ ----> inequality A

The solution of the inequality A is the shaded area below the dashed line $y=5-2x$

$x+5y>-7$ ---->inequality B

The solution of the inequality B is the shaded area above the dashed line $x+5y=-7$

The solution of the system of inequalities is the triangular shaded area between the two dashed lines

If a ordered pair is a solution of the system of inequalities, then the ordered pair must lie on the shaded area

Two points in the solution are(1,2) and (2,-1)see the attached figure

4 0

4 years ago

the value of machine depreciates every year by 5%. If the present value of the machine is 100000,what will be its value after 2

valentina_108 [34]

It starts out being worth $100,000 (One hundred thousand dollars). Every year it depreciates by 5% (0.05). What is the value after two years? We can find this by multiplying.

100,000 * 0.05 = 5,000
100,000 - 5,000 = 95,000

95,000 * 0.05 = 4,750
95,000 - 4,750 = 90,250

The machine, after two years, will be worth $90,250.

3 0

3 years ago

Read 2 more answers

1. Let the test statistics Z have a standard normal distribution when H0 is true. Find the p-value for each of the following sit

Nonamiya [84]

Answer:

1 a $p -value = 0.030054$

1b $p -value = 0.0029798$

1c $p -value = 0.0039768$

2a $p-value = 0.00099966$

2b $p-value = 0.00999706$

2c $p-value = 0.0654412$

Step-by-step explanation:

Considering question a

The alternative hypothesis is H1:μ>μ0

The test statistics is z =1.88

Generally from the z-table the probability of z =1.88 for a right tailed test is

$p -value = P(Z > 1.88) = 0.030054$

Considering question b

The alternative hypothesis is H1:μ<μ0

The test statistics is z=−2.75

Generally from the z-table the probability of z=−2.75 for a left tailed test is

$p -value = P(Z < -2.75) = 0.0029798$

Considering question c

The alternative hypothesis is H1:μ≠μ0

The test statistics is z=2.88

Generally from the z-table the probability of z=2.88 for a right tailed test is

$p -value = P(Z >2.88) = 0.0019884$

Generally the p-value for the two-tailed test is

$p -value = 2 * P(Z >2.88) = 2 * 0.0019884$

=> $p -value = 0.0039768$

Considering question 2a

The alternative hypothesis is H1:μ>μ0

The sample size is n=16

The test statistic is t = 3.733

Generally the degree of freedom is mathematically represented as

$df = n - 1$

=> $df = 16 - 1$

=> $df = 15$

Generally from the t distribution table the probability of t = 3.733 at a degree of freedom of $df = 15$ for a right tailed test is

$p-value = t_{3.733 , 15} = 0.00099966$

Considering question 2b

The alternative hypothesis is H1:μ<μ0

The degree of freedom is df=23

The test statistic is ,t= −2.500

Generally from the t distribution table the probability of t= −2.500 at a degree of freedom of df=23 for a left tailed test is

$p-value = t_{-2.500 , 23} = 0.00999706$

Considering question 2c

The alternative hypothesis is H1:μ≠μ0

The sample size is n= 7

The test statistic is ,t= −2.2500

Generally the degree of freedom is mathematically represented as

$df = n - 1$

=> $df = 7 - 1$

=> $df = 6$

Generally from the t distribution table the probability of t= −2.2500 at a degree of freedom of $df = 6$ for a left tailed test is

$t_{-2.2500 , 6} = 0.03272060$

Generally the p-value for t= −2.2500 for a two tailed test is

$p-value = 2 * 0.03272060 = 0.0654412$

4 0

3 years ago

509 to nearest hundreds

vladimir1956 [14]

509 rounded to the nearest hundred would be 500

6 0

4 years ago

Read 2 more answers

Can someone please help me?

Elenna [48]

Answer:

wagon wheel

radius = 1.5 ft

diameter = 3 ft

circumference = 9.4 ft

airplane propeller

radius = 24 in

diameter = 48 in

circumference = 150.7 in

Orange slice

radius = 3.185 cm

diameter = 6.37 cm

circumference = 20 cm

3 0

3 years ago