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3 years ago

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The width of a rectangle is 8 inches shorter than its length. The perimeter of the rectangle is 18 inches. What are the length a

nd width? Which equation models this problem? A. 8 • l = 18 – 8 B. 2l + 2(l – 8) = 18 C. l + (8 + 2l) = 18 D. l(l – 8) = 18

1 answer:

VashaNatasha [74]3 years ago

6 0

Answer:

B

Step-by-step explanation:

Let length be l

Width, w=l-8

Perimeter = 2(l+w) = 2l+2w

2l+2(l-8)=18

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Step-by-step explanation:

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3 years ago

Which function has a vertex at the origin?

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2 years ago

At an airport, 76% of recent flights have arrived on time. A sample of 11 flights is studied. Find the probability that no more

I am Lyosha [343]

Answer:

The probability is $P( X \le 4 ) = 0.0054$

Step-by-step explanation:

From the question we are told that

The percentage that are on time is p = 0.76

The sample size is n = 11

Generally the percentage that are not on time is

$q = 1- p$

$q = 1- 0.76$

$q = 0.24$

The probability that no more than 4 of them were on time is mathematically represented as

$P( X \le 4 ) = P(1 ) + P(2) + P(3) + P(4)$

=> $P( X \le 4 ) = \left n } \atop {}} \right.C_1 p^{1} q^{n- 1} + \left n } \atop {}} \right.C_2p^{2} q^{n- 2} + \left n } \atop {}} \right.C_3 p^{3} q^{n- 3} + \left n } \atop {}} \right.C_4 p^{4} q^{n- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{11- 1} + \left 11 } \atop {}} \right.C_2p^{2} q^{11- 2} + \left 11 } \atop {}} \right.C_3 p^{3} q^{11- 3} + \left 11 } \atop {}} \right.C_4 p^{4} q^{11- 4}$

$P( X \le 4 ) = \left 11 } \atop {}} \right.C_1 p^{1} q^{10} + \left 11 } \atop {}} \right.C_2p^{2} q^{9} + \left 11 } \atop {}} \right.C_3 p^{3} q^{8} + \left 11 } \atop {}} \right.C_4 p^{4} q^{7}$

$= \frac{11! }{ 10! 1!} (0.76)^{1} (0.24)^{10} + \frac{11!}{9! 2!} (0.76)^2 (0.24)^{9} + \frac{11!}{8! 3!} (0.76)^{3} (0.24)^{8} + \frac{11!}{7!4!} (0.76)^{4} (0.24)^{7}$

$P( X \le 4 ) = 0.0054$

4 0

3 years ago

B) Mrs. Shakya sold a jewellery at a loss of 5%. If she had sold it at Rs 5,200 more,

Nadusha1986 [10]

Answer:

$\huge{ \fbox{ \sf{Rs \: 40000}}}$

Step-by-step explanation:

$\sf{Let \: CP \: ( \: cost \: price) \: be \: x}$

$\sf{Loss \: \% \: = 5 \% \: }$

$\sf{Selling \: price = CP \: - \: loss \: \% \: of \: CP}$

$\mapsto{ \sf x - \frac{5}{100} \times x}$

$\mapsto{ \sf{x - \frac{x}{20} }}$

$\mapsto{ \sf{ \frac{x \times 20 - x}{20} }}$

$\mapsto{ \sf{ \frac{20x - x}{20}}}$

$\mapsto{ \sf{ \frac{19x}{20}}}$

Now, Finding the New selling price

$\sf{New \: SP \: ( \: selling \: price)} = \frac{19x}{20} + 5200$

$\mapsto{ \sf{ \frac{19x + 5200 \times 20}{20}}}$

$\mapsto{ \sf{ \frac{19x + 10400}{20}}}$

Finally, finding the Cost price :

We have, Profit % = 8 %

$\sf{Cost \: price = Selling \: price \: - \: P\% \: of \: CP}$

$\mapsto{ \sf{x = \frac{19x + 104000}{20} - 8\% \: of \: x}}$

$\mapsto{ \sf{x = \frac{19x + 104000}{20} - \frac{8x}{100} }}$

$\mapsto{ \sf{x = \frac{19x + 104000}{20} - \frac{2x}{25} }}$

$\mapsto{ \sf{x = \frac{5(19x + 104000) - 2x \times 4}{100} }}$

$\mapsto{ \sf{x = \frac{95x + 520000 - 8x}{100} }}$

$\mapsto{ \sf{x = \frac{87x + 520000}{100}}}$

$\mapsto{ \sf{100x = 87x + 520000}} \: \: ( \sf{Cross \: multiplication}$

$\mapsto{ \sf{100x - 87x = 520000}}$

$\mapsto{ \sf{13x = 520000}}$

$\mapsto{ \sf{x = \frac{520000}{13}}}$

$\mapsto{ \boxed {\sf{x = 40000}}}$

Therefore, the cost price of the jewellery is Rs 40000

Hope I helped!

Best regards! :D

~ $\sf{TheAnimeGirl}$

5 0

3 years ago