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oksano4ka [1.4K]

1 year ago

9

Fiona wrote the linear equation y = x – 5. When Henry wrote his equation, they discovered that his equation had all the same sol

utions as Fiona’s. Which equation could be Henry’s?

2 answers:

Oliga [24]1 year ago

7 0

His equation could be written in quadratic form, which is ax^2+bx=c

AVprozaik [17]1 year ago

6 0

Answer with Step-by-step explanation:

Fiona wrote the linear equation y = x – 5.

Henry wrote his equation, they discovered that his equation had all the same solutions as Fiona’s.

Henry's equation could be any equation which is a constant multiple of Fiona's equation.

i.e. it could be

2y=2(x-5)

i.e. 2y=2x-10

Hence, Henry's equation could be:

2y=2x-10

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1. Miguel is playing a game in which a box contains four chips with numbers written on them. Two of the chips have the number 1,

zzz [600]

Answer:

Step-by-step explanation:

Given that Miguel is playing a game

The box contains 4 chips, 2 with number 1, and other two differntly numbered as 3 and 5.

OUt of these 4, 2 chips are drawn

P(drawing same number) = 2C2/4C2 = $\frac{1}{6}$

Prob (drawing differnt numbers) = 1-1/6 = $\frac{5}{6}$

Hence prob of winning 2 dollars = $\frac{1}{6}$

Prob of losing 1 dollar = $\frac{5}{6}$

b) Expected value = sum of prob x amount won

= $\frac{1}{6}2+\frac{5}{6}(-1)=-\frac{1}{2}$

c) Miguel can expect to lose 1/2 dollars for every game he plays

d) If it is to be a fair game expected value =0

i.e. let the amount assigned be s

Then $\frac{1}{6}s-\frac{5}{6}=0\\s=5$

6 0

1 year ago

Read 2 more answers

Among a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in

Hitman42 [59]

Answer:

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval

(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

Step-by-step explanation:

We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of Americans who decide to not go to college = 48%

n = sample of American adults = 331

p = population proportion of Americans who decide to not go to

college because they cannot afford it

<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level

of significance are -1.645 & 1.645}

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\hat p-p$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

<u>90% confidence interval for p</u> = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

= [ $0.48 -1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ , $0.48 +1.96 \times {\sqrt{\frac{0.48(1-0.48)}{331} } }$ ]

= [0.4348, 0.5252]

(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].

(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.

3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.

So, the margin of error = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$

$0.015 = 1.645 \times \sqrt{\frac{0.48(1-0.48)}{n} }$

$\sqrt{n} = \frac{1.645 \times \sqrt{0.48 \times 0.52} }{0.015}$

$\sqrt{n}$ = 54.79

n = $54.79^{2}$

n = 3001.88 ≈ 3002

Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.

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In your problem:
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n = 130

The standard error can be calculated by the formula:
SE = √[p · (1 - p) / n]
= √[0.183 · (1 - 0.183) / 130]
= 0.0339

The standard error of the proportion is 0.034.

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2 years ago

A student wants to know how far above the ground the top of a leaning flagpole is. At high noon, when the sun is almost directl

MariettaO [177]

Answer:

15.43 ft

Step-by-step explanation:

To solve this question, one should use the concept of similar triangles.

The pole's shadow and the distance to the ground (x) form a similar triangle to the distance of the plumb bob from the base and the length of the plumb bob, respectively. Note that we do not need to know the measurements of the third side of the triangles to solve the problem. Therefore, the distance of the top of the pole to the ground is:

$14 in = 1.1667 ft$

$1.1667x=6*3\\x = 15.43$

The distance of the top of the pole to the ground is 15.43 ft.

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1 year ago

Aramis is adjusting a satellite because he finds it is not focusing the incoming radio waves perfectly. The shape of his satelli

Alika [10]

Given the equation of the parabola

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$2p=3,\\ \\p=1.5.$

The coordinates of the parabola focus are

$\left(x_0,y_0+\dfrac{p}{2}\right).$

Therefore, the focus is placed at point (4,3,75).

Answer: option D, 0.75 in. above the vertex

3 0

2 years ago

Read 2 more answers