4g+1 5 − G−7 4 = 2g−5 10

Malorie can work 10 math problems in 3 minutes. At this rate, how many problems can she work in one hour?

lisov135 [29]

D
If you multiply 3 by 10 and add a zero onto the ten in the question that is 100 in half an hour, all you need to do then is multiply that answer by two to get your answer.

4 0

4 years ago

Please help me out with this quiz! Alegbra One

Taya2010 [7]

Answer:

1B

2A

3C

4C

5C

6A

7A

8D

9B

10C

6 0

3 years ago

a private school admits no more than 100 students every year. additionally, at least 30 of these students must be girls, and the

Thepotemich [5.8K]

2/2 part of the equation is ≤ B
maybe 30/100 ≤ B

7 0

3 years ago

A company that produces computers recently tested the battery in its latest laptop in six separate trials. The battery lasted 8.

Anna35 [415]

confidence interval for the mean battery life in the new model is [7.9489, 8.3045].

What is confidence interval ?

A confidence interval, in statistics, refers to the chance that a population parameter can fall between a collection of values for an exact proportion of times.

Main body:

Formula for confidence interval is =

CI = x- bar ± z*s/√n where,

CI = confidence interval

x- bar = sample mean

z = confidence level value

{s} = sample standard deviation

{n} = sample size

given ;

n = 6

mean = (8.23+7.89+8.14+8.25+8.30+ 7.95)/6

= 8.127

value of z for 95% C.I. = 1.96

C.I. = 8.127 ± 1.96 * 0.22/√6

C.I. = 8.127 ± 0.1781

C.I. =[7.9490, 8.3044]

Hence correct option is A.

To know more about confidence interval , visit:;

brainly.com/question/15712887

#SPJ4

8 0

1 year ago

A certain square is to be drawn on a coordinate plane. One of the vertices must be on the origin, and the square is to have an a

Scrat [10]

Answer:

The answer is (C) 8

Step-by-step explanation:

First, let's calculate the length of the side of the square.

$A_{square}=a^2$ , where $a$ is the length of the side. Now, let's try to build the square. First we need to find a point which distance from (0, 0) is 10. For this, we can use the distance formula in the plane:

$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ which for $x_1=0$ and $y_1 = 0$ transforms as $d=\sqrt{(x_2)^2 + (y_2)^2}$ . The first point we are looking for is connected to the origin and therefore, its components will form a right triangle in which, the Pythagoras theorem holds, see the first attached figure. Then, $x_2$ , $y_2$ and 10 are a Pythagorean triple. From this, $x_2= 6$ or $x_2=8$ while $y_2= 6$ or $y_2=8$ . This leads us with the set of coordinates:

$(\pm 6, \pm 8)$ and $(\pm 8, \pm 6)$ . (A)

The next step is to find the coordinates of points that lie on lines which are perpendicular to the lines that joins the origin of the coordinate system with the set of points given in (A):

Let's do this for the point (6, 8).

The equation of the line that join the point (6, 8) with the origin (0, 0) has the equation $y = mx +n$ , however, we only need to find its slope in order to find a perpendicular line to it. Thus,

$m = \frac{y_2-y_1}{x_2-x_1} \\m = \frac{8-0}{6-0} \\m = 8/6$

Then, a perpendicular line has an slope $m_{\bot} = -\frac{1}{m} = -\frac{6}{8}$ (perpendicularity condition of two lines). With the equation of the slope of the perpendicular line and the given point (6, 8), together with the equation of the distance we can form a system of equations to find the coordinates of two points that lie on this perpendicular line.

$m_{\bot}=\frac{6}{8} = \frac{8-y}{6-x}\\ 6(6-x)+8(8-y)=0$ (1)

$d^2 = \sqrt{(y_o-y)^2+(x_o-x)^2} \\(10)^2=\sqrt{(8-y)^2+(6-x)^2}\\100 = \sqrt{(8-y)^2+(6-x)^2}$ (2)

This system has solutions in the coordinates (-2, 14) and (14, 2). Until here, we have three vertices of the square. Let's now find the fourth one in the same way we found the third one using the point (14,2). A line perpendicular to the line that joins the point (6, 8) and (14, 2) has an slope $m = 8/6$ based on the perpendicularity condition. Thus, we can form the system:

$\frac{8}{6} =\frac{2-y}{14-x} \\8(14-x) - 6(2-y) = 0$ (1)

$100 = \sqrt{(14-x)^2+(2-y)^2}$ (2)

with solution the coordinates (8, -6) and (20, 10). If you draw a line joining the coordinates (0, 0), (6, 8), (14, 2) and (8, -6) you will get one of the squares that fulfill the conditions of the problem. By repeating this process with the coordinates in (A), the following squares are found:

(0, 0), (6, 8), (14, 2), (8, -6)

(0, 0), (8, 6), (14, -2), (6, -8)
(0, 0), (-6, 8), (-14, 2), (-8, -6)
(0, 0), (-8, 6), (-14, -2), (-6, -8)

Now, notice that the equation of distance between the two points separated a distance of 10 has the trivial solution $(\pm10, 0)$ and $(0, \pm10)$ . By combining this points we get the following squares:

(0, 0), (10, 0), (10, 10), (0, 10)
(0, 0), (0, 10), (-10, 10), (-10, 0)
(0, 0), (-10, 0), (-10, -10), (0, -10)
(0, 0), (0, -10), (-10, -10), (10, 0)

See the attached second attached figure. Therefore, 8 squares can be drawn

8 0

3 years ago