In the past, the mean running time for a certain type of flashlight battery has been 9.7 hours. The manufacturer has introduced a change in the production method and wants to perform a hypothesis test to determine whether the mean running time has increased as a result. The hypotheses are: H 0: μ = 9.7 hours H a: μ > 9.7 hours where μ is the mean running time of the new batteries . Explain the meaning of a Type I error.

So you want to go to your calculator type 42 divided by zero, and you should get a 52.5.

Answer:

clockwise

Q= (5,-2)

R= (-1,-3)

S= (0,0)

T= (3,-2)

counterclockwise

Q= (-5,2)

R= (1,3)

S= (0,0)

T= (-3,-2)

Step-by-step explanation:

this might not be right im a little rusty on my transformations but here goes nothing.

i dont know if this is a clockwise or a counter clockwise rotation so ill do both options.

clockwise (x,y) - (y,-x)

q=(2,5) - (5,-2)

r=(3,-1) - (-1,-3)

s=(0,0) - (0,0)

t=(-2,3) - (3,2)

counterclockwise (x,y) - (-y,x)

q=(2,5) - (-5,2)

r=(3,-1) - (1,3)

s=(0,0) - (0,0)

t=(-2,3) - (-3,-2)

Answer:

The answer is -4/21

Step-by-step explanation:

Answer:

50 employees

Step-by-step explanation:

The total number of employees the business has will have to be made up the full-time employees and part-time employees.

The number of full-time employees = 15

Let the total number of  employees = x

[tex]15 = 30 percent of x[/tex]

[tex]15 = 30/100 \times x\\x=1500/30 = 50[/tex]

The company has a total of 50 employees

Answer:

What's the value of 6.7×0.54

=

3.618

Step-by-step explanation:

Answer:

3.618

Step-by-step explanation:

3.618

Answer:

25

Step-by-step explanation:

Answer:

I think its B

Step-by-step explanation:

I hope it helps

Answer:

the answer is -28/20 hope it helps

Answer:

-14

Step-by-step explanation

Answer:

[tex]x=\frac{3\pi}{4}\\\\x=\frac{5\pi}{4}[/tex]

Step-by-step explanation:

Trigonometric Equations

It's required to solve:

[tex]\cos x+\sqrt{2}=-\cos x[/tex]

for [tex]x\in [0,2\pi)[/tex]

Adding cos x:

[tex]2\cos x+\sqrt{2}=0[/tex]

Subtracting [tex]\sqrt{2}[/tex]

[tex]2\cos x=-\sqrt{2}[/tex]

Dividing by 2:

[tex]\displaystyle \cos x=-\frac{\sqrt{2}}{2}[/tex]

Solving for x:

[tex]\displaystyle x=\arccos\left(-\frac{\sqrt{2}}{2}\right)[/tex]

We need to find the angles whose cosine is [tex]-\frac{\sqrt{2}}{2}[/tex] over the given interval.

These angles lie on the quadrants III and IV respectively and they are:

x=135°, x=225°

Converting to radians:

135 * π / 180 = 3π/4

225 * π / 180 = 5π/4

The two solutions are:

[tex]\mathbf{x=\frac{3\pi}{4}}[/tex]

[tex]\mathbf{x=\frac{5\pi}{4}}[/tex]

Answer:

The correct answer is option C

Step-by-step explanation:

Answer:

3x+48

add 3+45=48

Step-by-step explanation:

Hope this helps! Please consider marking brainliest! I hope you have an amazing day. Always remember, your smart and you got this! -Alycia :)

Answer:

3(x+16)

Step-by-step explanation:

45+3=48

3x/3=x

48/3=16

3(x+16)

Answer:

so one thing you can do is to put the standard form into slope-intercept form

so 6x+y= -4 into y=-6x -4

now, you put a point on (0,-4)

(x,y)

now the slope is -6/1 so you rise (go down) -6 points and run (sideways) 1

I did 40+32 in got 72

Or 40-32=8

Answer:

18

Step-by-step explanation:

Step 1)

54 ÷ 3 = 18

Answer:

54/3

Step-by-step explanation:

Answer:

100

Step-by-step explanation:

20% of 500 is 100

Answer:

$100

Step-by-step explanation:

20% is the same as 1/5. 1/5 of 500 is 100. Then add dollar signs, and you have your answer.

Answer:

48.75

Step-by-step explanation:

Answer:

Task

Kipton has a digital scale. He puts a marshmallow on the scale and it reads 7.2 grams. How much would you expect 10 marshmallows to weigh? Why?

Kipton takes the marshmallows off the scale. He then puts on 10 jellybeans and then scale reads 12.0 grams. How much would you expect 1 jellybean to weigh? Why?

Kipton then takes off the jellybeans and puts on 10 brand-new pink erasers. The scale reads 312.4 grams. How much would you expect 1,000 pink erasers to weigh? Why?

IM Commentary

The purpose of this task is to help students

Recognize that in a multi-digit number, a digit in one place represents 10 times as much as it represents in the place to its right and 1/10 of what it represents in the place to its left (5.NBT.1)

By setting the task in the context of weighing objects and bundles of 10, 100, and 1,000 objects, it helps students visualize that bundling 10 units of a given place value will create 1 unit of the next highest place value. For example, taking 10 of an item that weighs 4.2 grams will result in 42 grams because it is 10 groups of 4 grams that weights 40 grams all together, and 10 groups of 0.2 grams that weigh 2 grams all together. The task allows students to explore both the structure of our place value system and how we use that to efficiently multiply and divide powers of 10.

Though this task is written as questions to be discussed or answered by students, the parts of this task are most valuable as a set of scenarios to be infused throughout a unit on place value. The teacher can actually bring in a digital scale and go through a series of explorations with students. It might be natural to start with students weighing one object and making predictions about how much 10 or 100 of these objects would weigh. From there, the teacher may want to have students start weighing sets of 10 or 100 objects and work backwards to think about the weight of 1 object.

The digital scale has several advantages: the weight will always appear in decimal form, thereby making it perfect for students to start reasoning about shifts in decimal place value from a more intuitive place. Moreover, the digital scale will not always show the expected answer. For example, something that weighs 3.5 grams alone might weigh 35.4 grams (rather than 35 grams) when taken as a group of 10. This slight difference provides an excellent opportunity to talk about rounding error. It also requires students to think beyond the rules of sliding a decimal point to the right when multiplying by powers of 10.

If students were still developing the rules for multiplying and dividing by powers of 10, this task would also incorporate MP8, Look for and express regularity in repeated reasoning. It also incorporates parts of MP6, Attend to precision, in that a higher-level discussion will push students to reason why one marshmallow might weigh 7.2 grams, but 10 marshmallows might weigh 7.19 grams. See the solution for part (a) for further elaboration.

Solution

Solution:

10 marshmallows should weigh 72 grams. Students might use repeated addition, multiplication or reason that each digit’s place value will be multiplied by a factor of 10:

10×(7+0.2)(10×7)+(10×0.2)70+272===

If students have had practice using digital scales in class, some students might respond with guesses that are close such as 73 grams or 72.4 grams because the original 7.2 grams may have been rounded to the nearest tenth. If, for example, the original actual weight were 7.24 grams, then the weight of ten marshmallows on a scale that rounds to the nearest tenth of a gram would be 72.4 grams.

1 jellybean should weigh 1.2 grams. Students may come to the solution a number of ways.

Students might notice that dividing 12 by 10 could be expressed in fraction form 1210. Students might further reason that

1210=1010+210

or 1.2.

Students might alternately reason that 12÷10 is the same as

(10÷10)+(2÷10)=1+0.2=1.2

Students might also use the rule that they are beginning to develop about sliding the decimal point one place to the left to divide by 10.

1,000 pink erasers should weigh about 31,240 grams. Students may come to the solution a number of ways:

A student might reason that first, it would be necessary to find out what 100 erasers would weigh by multiplying by a factor of 10 and then further multiplying by another factor of 10 to find out what 1,000 erasers would weigh. This is represented by

312.4×10×10

In this solution method, the student reasoned that multiplying by a factor of 10 and then another factor of 10 would be the same as multiplying by a factor of 100:

312.4×10×10312.4×100100×(300+10+2+0.4)30,000+1,000+200+4031,240====

Another student might want to find the weight of a single eraser. If 10 erasers weigh 312.4 grams, then one eraser must weigh 31.24 grams. From there the student could multiply the weight of one eraser by a factor of 1,000:

1000×(30+1+0.22+0.04)30,000+1,000+200+4031,240==

Step-by-step explanation:

You would expect 1,000 pink erasers to weigh 31,240 grams.

If 10 erasers weigh 312.4 grams, and to get 1,000 you multiply by 100, if you multiply 312.4 by 100 you get 31240.

Or if you use rates

[tex]\frac{312.4}{10} =\frac{31240}{1000}[/tex]

(when you multiply by 100)

Answer:

it's the first answer

Step-by-step explanation:

literally every other answer is ruled out because D is the greatest variable and can't be smaller than anything else

Answer:

0.6

Step-by-step explanation:

Brainliest please

Answer: The slope is 4/3

Step-by-step explanation:

Whatever number is attatched to the "x" is the slope.

Answer:

Slope: M = 4/3

Y-Intercept: : (0,−1)

Step-by-step explanation:

Answer: B

Step-by-step explanation: Because the positive numbers changed to negatives meaning that it flipped over an axis and since both of them changed it would be the y-axis

Answer:

sure so quadrant one is top right and its positive positive and quadrant 2 top left and is negative positive quadrant 3 is bottom left and negative negative quadrant 4 is bottom right and its positive negative

Step-by-step explanation:

Answer:

Slope = 4.000/2.000 = 2.000

Slope = 2.000/2.000 = 1.000

Step-by-step explanation:

the first answer is for the first problem and the second answer is for the second problem. Hoped it helped i tried my best.

Answer:

x=u+k

Step-by-step explanation:

Answer:

x = k+u

Step-by-step explanation:

u = x - k

+k       +k

u + k = x

Hope this helps!

Given:

Point is T(-3,8).

To find:

The coordinates of T' after [tex]R(y-axis)\circ R(x-axis)[/tex].

Solution:

We know that, [tex]R(y-axis)\circ R(x-axis)[/tex] means the figure reflected across the x-axis then reflected across y-axis.

If a figure reflected across x-axis, then

[tex](x,y)\to (x,-y)[/tex]

[tex]T(-3,8)\to T_1(-3,-8)[/tex]

If a figure reflected across y-axis, then

[tex](x,y)\to (-x,y)[/tex]

[tex]T_1(-3,-8)\to T'(-(-3),-8)[/tex]

[tex]T_1(-3,-8)\to T'(3,-8)[/tex]

Therefore, the required point is T'(3,-8).

Answer:

One solution which is 8

Step-by-step explanation:

Hope this helps. Pls give brainliest.

Answer:

80, 95, 110, 125, 140

Step-by-step explanation: