A bicyclist rides 1/5 mile in 1/65 hour. Write this rate as a unit rate.

Factor to find the zeros of the function defined by the quadratic expression. −13x2 − 130x − 273

rewona [7]

Answer The Zeros are x= -3,-7

7 0

2 years ago

Researchers doing a study comparing time spent on social media and time spent on studying randomly sampled 200 students at a maj

Charra [1.4K]

Answer:

d. 0.0023

Step-by-step explanation:

Calculation for the probability of getting a sample average of 1.8 or less

Using this formula

z score =(X-mean)/Standard errror

Where,

Standard error =Standard deviation/√(n)

Let plug in the formula

P(Xbar<1.8)=P(Z<(1.8 hours per day -2.2 hours per day)/(2 hours /√(200))

P(Xbar<1.8)=P(Z<-0.4/(2/14.14))

=P(Z<-0.4/0.1414)

=P(Z<-2.83)

Now let find the Z score of P(Z<-2.83) using the z table

P=0.0023

Therefore the probability of getting a sample average of 1.8 or less is:0.0023

5 0

2 years ago

vani is building a rectangular play area. The length of the play area is 7.5 meters. The width of the play area is 5.3 meters. I

horsena [70]

39.75 square meters

5 0

3 years ago

HELP ASAP PLZZZZ

Tcecarenko [31]

QUESTION 1

The given system of equations is

$3d - e = 7...eqn(1)$
$d + e = 5...eqn(2)$

To solve by linear combination, we add equation (1) to equation (2) to get,

$3d + d= 7 + 5$

$4d = 12$

We divide through by 4 to obtain,

$d = \frac{12}{4}$

$d = 3$

We put d=3 into equation (2) to get,

$3+ e = 5$

$e = 5 - 3$

$e = 2$

$\boxed {The \: solution \: is \: (3, 2)}$

QUESTION 2

The given system is

4x + y = 5 ...eqn(1)

3x + y = 3 ...eqn(2)

To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.

This will give us,

$4x - 3x = 5 - 3$

This implies that,

$x = 2$

Put x=3 into equation (1) to get,

$4(2) + y = 5$

$8+ y = 5$

$y = 5 - 8$

$y = - 3$

The solution is

$(2,-3)$

QUESTION 3

We want to solve the system;

a – 2b = –2 ....eqn(1)

2a + 2b = 14...eqn(2)

by linear combination.

We need to add equation (1) to equation (2) to eliminate b.

This implies that,

$2a + a = 14 + - 2$

Simplify,

$3a = 12$

Divide both sides by 3 to get,

$a = 4$
Put a=4 into equation (2) to obtain,

$2(4) + 2b = 14$

$8 + 2b = 14$
$2b = 14 - 8$

$2b = 6$

$b = 3$

The ordered pair in the form (a, b) is

$(4,3)$

QUESTION 4

The given system of equations is

11x + 4y = 18 ...eqn(1)

3x + 4y = 2 ...eqn(2)

We subtract equation (2) from equation (1) to get,

$11x - 3x = 18 - 2$

$8x = 16$

$x = 2$

Put x=2 into equation (2) to obtain,

$3(2) + 4y = 2$

This implies that,

$6 + 4y = 2$

$4y = 2 - 6$

$4y = - 4$

$y=-1$

The correct answer is (2,-1).

QUESTION 5

The given system is ;

2d + e = 8...eqn1

d – e = 4...eqn2

We add the two equations to eliminate e.

This implies that,

$2d + d = 8 + 4$

$3d = 12$

We divide both sides by 3 to get,

$d = 4$

We put d=4 into equation (2) to get,

$4 - e = 4$

$- e = 4 - 4$

$- e = 0$

$e = 0$

The solution is

$(4,0)$

7 0

3 years ago

(3a2 + 2a - 2) - (a2 - 3a + 7)

drek231 [11]

Remember to distribute invibile -1
simlify 2nd part
-(a^2-3a+7)=-1(a^2-3a+7)=-a^2+3a-7

now we have
3a^2+2a-2-a^2+3a-7=
3a^2-a^2+2a+3a-2-7=
2a^2+5a-9

3 0

3 years ago