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IrinaVladis [17]

2 years ago

6

A teacher gave her class two exams; 60% of the class passed the second exam, but only 48% of the class passed both exams. What p

ercent of those who passed the second exam also passed the first exam?

2 answers:

zubka84 [21]2 years ago

7 0

I think 80%

I also got stuck on it and searched it up I think its 80% but if I'm wrong I'm sorry

Dahasolnce [82]2 years ago

6 0

Answer:

80%

Step-by-step explanation:

If 60% of the class passed the second exam and only 48% passed the first your solution would have to be of the 48%. Therefore 0.48/0.60=0.8/1.0=80%

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C2 = 39<br><br> (c squared) <br><br> please help :(

iVinArrow [24]

<em>Answer:</em>

<em></em> $\boxed{c = \sqrt{39} / c= - \sqrt{39} }$ <em></em>

<em>This " / " means or, btw!</em>

<em>Step-by-step explanation:</em>

<em></em> $c^2 = 39$ <em></em>

<em>⇒Take square root.</em>

<em></em> $c =$ <em> ± </em> $\sqrt{39}$ <em></em>

<em></em> $c = \sqrt{39}$ <em> or </em> $c = -\sqrt{39}$ <em></em>

<em>Hope this helped you!</em>

7 0

2 years ago

: A personal computer manufacturer buys 36% of its chips from Japan and the rest from the United St...

meriva

Answer: 0.9862

Step-by-step explanation:

Given : The probability that the chips belongs to Japan: P(J)= 0.36

The probability that the chips belongs to United States : P(U)= 1-0.36=0.64

The proportion of Japanese chips are defective : P(D|J)=0.017

The proportion of American chips are defective : P(D|U)=0.012

Using law of total probability , we have

$P(D)=P(D|J)\times P(J)+P(D|U)\times P(U)\\\\\Rightarrow\ P(D)=0.017\times0.36+0.012\times0.64=0.0138$

Thus , the probability that chip is defective = 0.0138

Then , the probability that a chip is defect-free= $1-0.0138=0.9862$

3 0

2 years ago

A triangle has side lengths 4 inches 5 inches and 7 inches. is the triangle a right triangle

garri49 [273]

Hey there!

Yes this triangle is a right triangle. Leg lengths 4 and 5 in. would be the legs and the 7 inch would be the hypotenuse.

Hope this helped!

3 0

2 years ago

Beginning in 2000, an orchestra increased its ticket price by a constant amount each year until 2010.

Scilla [17]

Answer:

$35.50

Step-by-step explanation:

55.50-48= 7.5

2008-2005 = 3

7.5 divided by 3 = 2.5 per year

2005-2000 = 5 multipled by 2.5 = 12.5

48-12.5=35.5

6 0

1 year ago

Factorize 2a^2+7a-15

loris [4]

$\huge\bf \pink {\underline{Solution :-}}$

$2 {a}^{2} + 7a - 15$

$= 2 {a}^{2} + 10a - 3a - 15$

$= 2a(a + 5) - 3(a + 5)$

$= (a + 5)(2a - 3)$

$\sf \red{Hence, Answer \: is \: (a + 5)(2a - 3).}$

$\\$

$\bf \purple{ \underline{ Important \: Formulas \: for \: Factorization :-}}$

$• \: {a}^{2} + 2ab + {b}^{2} = {(a + b)}^{2}$

$• \: (a + b)(a - b) = {a}^{2} - {b}^{2}$

$• \: {a}^{2} - 2 ab + {b}^{2} = {(a - b)}^{2}$

$• \: {a}^{3} + 3 {a}^{2} b + 3a {b}^{2} + {b}^{3} = {(a + b)}^{3}$

$• \: {a}^{3} - 3 {a}^{2} b + 3a {b}^{2} - {b}^{3} = {(a - b)}^{3}$

$• \: {(a + b)}^{2} + {(a - b)}^{2} = 2( {a}^{2} + {b}^{2} )$

$• \: {(a + b)}^{2} - {(a - b)}^{2} = 4ab$

$• \: (a + b)( {a}^{2} -ab + {b}^{2} ) = {a}^{3} + {b}^{3}$

$• \: (a - b)( {a}^{2} + ab + {b}^{2} ) = {a}^{3} - {b}^{3}$

$• \: {( \frac{a + b}{2} )}^{2} - ( {\frac{a - b}{2} } )^{2} = ab$

5 0

2 years ago

Read 2 more answers