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inysia [295]
2 years ago
15

Find the sum. 20 + (-8)

Mathematics
1 answer:
Alenkasestr [34]2 years ago
3 0

Answer:

12

ok ndjdjdjdhdhfkfgkgkgk

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Hotdogs are sold in packages The graph below shows the total number of hotdogs sold relative to the number of packages what does
soldier1979 [14.2K]

Answer:

do you still need help

Step-by-step explanation:

to this question

5 0
2 years ago
Select the correct answer. Which statement about an electromagnet is true? A. It doesn’t function without a source of current. B
zubka84 [21]

Answer:I’m going to say it’s A I apologize if it’s wrong

Step-by-step explanation:

Magnets are magnetic fields that compares together when attracted close to one another so that’s Why I’m choosing A because it’s close to what I think

3 0
3 years ago
Before changes to its management staff, an automobile assembly line operation had a scheduled mean completion time of minutes. T
Sergio [31]

No, we don't have evidence to support that the mean completion time under new management has decreased but we can conclude that it remains at 15.5 minutes.

Given mean of 15.5 minutes , standard deviation of 1.7 minutes, sample size of 90 and sample mean of 15.4 minutes.

We can do the following study for conclusion:

Firstly the null hypothesis is

H_{0}: x=15.5

The alternate hypothesis is

H_{1}: x < 15.5

since the value is less than this is a one tailed test.

Z=x bar-x/d/\sqrt{n}

where x is sample mean and d is standard deviation.

Z=15.4-15.5/1.7/\sqrt{90}

=-0.1/1.7/9.4868

=-0.560

Critical value of Z at 0.1 level of significance

Z=-1.28

We fails to reject the null hypothesis since -0.560>-1,28

Hence we don't have evidence to support that the mean completion time under new management has decreased but we can conclude that it remains at 15.5 minutes.

Learn more about hypothesis at brainly.com/question/11555274

#SPJ4

Question is incomplete. It should include:

mean of 15.5 minutes ,

standard deviation of 1.7 minutes,

sample size of 90

and sample mean of 15.4 minutes.

5 0
1 year ago
Problem 1.1
Vesna [10]

Answer:

Step-by-step explanation:

1 table and 2 chairs (2-seat table)

1 table and 4 chairs (4-seat table)

120 people so we need 120 chairs

Some Possibilities  :

120 /4 = 30

0( 2-seat table) and 30 (4- seat tables) because 0·2 + 30·4 = 0+120 = 120

2( 2-seat table) and 29 (4- seat tables) because 2·2 + 29·4 = 4+ 116= 120

4( 2-seat table) and 28 (4- seat tables) because 4·2 + 28·4 = 8+ 112= 120

...

120/2 = 60

60( 2-seat table) and 0 (4- seat tables) because 60·2 + 0·4 = 120 + 0 = 120

58( 2-seat table) and 1 (4- seat tables) because 58·2 + 1·4 = 116 + 4 = 120

56( 2-seat table) and 2 (4- seat tables) because 56·2 + 2·4 = 112 + 8 = 120

...

5 0
2 years ago
Simplify
Diano4ka-milaya [45]

Answer:

\huge\boxed{\sqrt[4]{16a^{-12}}=2a^{-3}=\dfrac{2}{a^3}}

Step-by-step explanation:

16=2^4\\\\a^{-12}=a^{(-3)(4)}=\left(a^{-3}\right)^4\qquad\text{used}\ (a^n)^m=a^{nm}\\\\\sqrt[4]{16a^{-12}}=\bigg(16a^{-12}\bigg)^\frac{1}{4}\qquad\text{used}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\=\bigg(2^4(a^{-3})^4\bigg)^\frac{1}{4}\qquad\text{use}\ (ab)^n=a^nb^n\\\\=\bigg(2^4\bigg)^\frac{1}{4}\bigg[(a^{-3})^4\bigg]^\frac{1}{4}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=2^{(4)(\frac{1}{4})}(a^{-3})^{(4)(\frac{1}{4})}=2^1(a^{-3})^1=2a^{-3}\qquad\text{use}\ a^{-n}=\dfrac{1}{a^n}

=2\left(\dfrac{1}{a^3}\right)=\dfrac{2}{a^3}

5 0
3 years ago
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