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3 years ago

Hannah and Francine have $120. Hannah and Peter have $230. Peter has 6 times as much money as Francine. How much money does Hann

ah have?

Mathematics

1 answer:

aivan3 [116]3 years ago

5 0

$x-Francine's\ money\\6x-Peter's\ money\\120-x-Hanna's\ money\\\\120-x+6x=230\\-x+6x=230-120\\5x=110\ \ \ |:5\\x=22\\\\6x=6\cdot22=132\\\\120-22=98\\\\Hanna\ has\ \$98$

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Help,anyone can help me do quetion

a_sh-v [17]

Answer:

B1: x = 51

C1: x = 73, y = 34

Step-by-step explanation:

B1:

By exterior angle theorem:

x° + 54° = 105°

x° = 105° - 54°

x° = 51°

x = 51

x° = 73°(Alternate angles)

x = 73

y° = 34° (corresponding angles)

y = 34

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2 years ago

100 points pls help

blsea [12.9K]

Answer:

a) 2/3p
b) q + 1/2p

Step-by-step explanation:

Given

AB = p
AD = q

DC /AB = 3/2 ⇒
DC = 3/2AB = 3/2p

BC = AD + (DC - AB)
BC = q + (3/2p - p) = q + 1/2p

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3 years ago

Read 2 more answers

Avaluate 2^3x-1 for x=1

DENIUS [597]

Answer:

4 or 7 (Check the explanation to see which one your equation actually is.

Step-by-step explanation:

One thing I want to clarify for you, It's Evaluate, not Avaluate.

Okay so, we want to find the value of 2^3x-1 and we know x = 1.

You didn't really clarify if the expression was $2^{3x-1}$ or $2^{3x}-1$ , so I'll be doing both:

For $2^{3x-1}$ , we should plug in x to get $2^{3(1)-1}$ and then simplify to get $2^{2}$ .

2 to the power of 2 or $2^{2}$ is equal to 2 * 2 or 4.

The second one, $2^{3x}-1$ , we should plug in x to get $2^{3(1)}-1$ and then to become $2^{3}-1$ . 2 to the power of 3 or $2^{3}$ is 8 and then minus 1 is 7.

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3 years ago

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Please help question in the picture

tigry1 [53]

There really isn’t a possible answer since there is not enough information given. Not from what I have learned so far at least, but using common sense, and the process of elimination I would say 30

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2 years ago

A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean l

tankabanditka [31]

Answer:

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

Step-by-step explanation:

Step(i):-

Given that the mean of the Population = 95

Given that the standard deviation of the Population = 5

Let 'X' be the random variable in a normal distribution

Let X⁻ = 96.3

Given that the size 'n' = 84 monitors

Step(ii):-

The Empirical rule

$Z = \frac{x^{-} -mean}{\frac{S.D}{\sqrt{n} } }$

$Z = \frac{96.3 - 95}{\frac{5}{\sqrt{84} } }$

Z = 2.383

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = P(Z≥2.383)

= 1- P( Z<2.383)

= 1-( 0.5 -+A(2.38))

= 0.5 - A(2.38)

= 0.5 -0.4913

= 0.0087

Final answer:-

The probability that the mean monitor life would be greater than 96.3 months in a sample of 84 monitors

P(X⁻ ≥ 96.3) = 0.0087

3 0

2 years ago