Does anyone remember that one math game that had origami animals as characters. If so tell me!

From Environment Canada: "Our experts on weather and climate have combed through 45 years of weather records, and determined the

miskamm [114]

Hat grade are u? I'm just asking cause the formulas are going with grade I could help if u want me too

7 0

2 years ago

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PLEASE HELP!!! ASAP The function graphed approximates the height of a rock, in meters, x seconds after it falls from a cliff

nasty-shy [4]

The vertical numbers are the distance and the horizontal numbers are the time in seconds.

Look at where the line is located at the number 20.

It si between the 2 and the 3 on the horizontal line, but you can see it is closer to the 2 than it is the 3, so the time would be 2.2s

8 0

1 year ago

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Daniel deposited $500 into a savings account and after 8 years, his investment is worth $807.07. The equation A = d(1.005)12t mo

shtirl [24]

Answer:

Step-by-step explanation:

The equation A = d(1.005)^12t modelling the value of Daniel’s investment shows a monthly compounded interest. This means that the interest is compounded 12 times in a year.

We can confirm by inputting the given values

t = 8 years

d = 509

Therefore,

A = 500(1.005)12 × 8

A = 500(1.005)^96

A = $807.07

Therefore, the true statements are

Increases

Exponential

Never Decrease

5 0

2 years ago

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

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

So, 90% confidence interval for the population proportion, p is ;

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

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

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

= [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%.

5 0

2 years ago

Can some one please please help me? solve for x 5 /7x + 1/7 =63 enter your answer in the box below.

Soloha48 [4]

This is confusing but I think yo umean
$\frac{x+5}{7x} + \frac{1}{7} =63$
times both sides by 7x
x+5+x=441x
2x+5=441x
minus 5 both sides
2x=436
divide both sides by 2
x=218

4 0

1 year ago