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2 years ago

4. Який ти знаєш хімічний посуд?

1 answer:

6 0

Основна стаття для цієї категорії: Лабораторний посуд. Вікісховище має мультимедійні дані за темою: Лабораторний посуд. Сторінки в категорії «Лабораторний ...

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Nichole gets paid $13 and hour washing cars.

Alex

Answer:

A) y = 13x

B) Slope is 13, intercept is 0

Step-by-step explanation:

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What is the volume of the prism? ft³

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3 years ago

Can someone help me?

Lana71 [14]

Answer:

I'm not sure from my answer but I'm going to try

A = { a^2 - 4 = 0 }

A = { a^2 = 0+ 4 }

A = { a^2 = 4}

Take root of both sides

A = {a = 2}

A = {2}

B = { -2, -1, 0, 1, 2}

C = { 1, 2, 3, 4, 5, 6,7}

i) A\B

= {2} \ {-2,-1,0,1,2}

= ⌀ ( "null set" )

ii) (A\B) \C

= ( {2} \ {-2,-1,0,1,2} ) \ C

= ⌀ \ C

= ⌀ \ { 1, 2, 3, 4, 5, 6,7}

= ⌀ ( "null set" )

iii) (A ∪ B) \ C

= ( { 2 } ∪ { -2, -1, 0, 1, 2} ) \ C

= { -2, -1, 0, 1, 2} \C

= { -2, -1, 0, 1, 2} \ { 1, 2, 3, 4, 5, 6,7}

= { -2, -1, 0 }

iv) ( A\B) ∩ ( A\C)

= ( { 2 } \ {-2,-1,0,1,2} ) ∩ ( { 2 } \ { { 1, 2, 3, 4, 5, 6,7} )

= ⌀ ∩ ⌀

= ⌀

v) ( A ∩ B ) \ C

= ( { 2 } ∩ { -2, -1, 0, 1, 2 } ) \ C

= { 2 } \ { 1, 2, 3, 4, 5, 6, 7 }

= ⌀ "null set"

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3 years ago

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$2, because if you do 8*25 you get 200 and if you divide that by 100 you get a total of 2 making the tax on the purchase $2.

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3 years ago

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A cookie factory monitored the number of broken cookies per pack yesterday.

trapecia [35]

Answer:

Confidence Interval - 2.290 < S < 2.965

Step-by-step explanation:

Complete question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 50 cookies. The mean is 23.33 and the standard deviation is 2.6. Construct a 80% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies.

Solution

Given

n=50

x=23.33

s=2.6

Alpha = 1-0.80 = 0.20

X^2(a/2,n-1) = X^2(0.10, 49) = 63.17

sqrt(63.17) = 7.948

X^2(1 - a/2,n-1) = X^2(0.90, 49) = 37.69

sqrt(37.69) = 6.139

s*sqrt(n-1) = 18.2

$s\sqrt{\frac{n-1}{X^2 _{(n-1), \frac{\alpha }{2} } }$ $\leq \sigma \leq s\sqrt{\frac{n-1}{X^2 _{(n-1), 1-\frac{\alpha }{2} } }$

confidence interval:

(18.2/7.948) < S < (18.2/6.139)

2.290 < S < 2.965

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3 years ago