A dime has the same value of 10 pennies. Marley brought 290 pennies to the bank. How many dimes did Marley get?

Aubrey and Charlie are driving to a city that is 120 miles from their house. They have already traveled 20 miles, and they are d

mario62 [17]

Answer:

f(1.5) = _95_

The value indicates that after_1.5 hours_

hours of driving, Aubrey and Charlie will be_95_

miles from home.

Step-by-step explanation:

8 0

2 years ago

Which score has a higher relative position, a score of 271.2 on a test for which = 240 and , or a score of 63.6 on a test for wh

Fittoniya [83]

Answer:

The score of 271.2 on a test for which xbar = 240 and s = 24 has a higher relative position than a score of 63.6 on a test for which xbar = 60 and s = 6.

Step-by-step explanation:

Standardized score, z = (x - xbar)/s

xbar = mean, s = standard deviation.

For the first test, x = 271.2, xbar = 240, s = 24

z = (271.2 - 240)/24 = 1.3

For the second test, x = 63.6, xbar = 60, s = 6

z = (63.6 - 60)/6 = 0.6

The standardized score for the first test is more than double of the second test, hence, the score from the first test has the higher relative position.

Hope this Helps!!!

3 0

1 year ago

On a coordinate plane, square P Q R S is shown. Point P is at (4, 2), point Q is at (8, 5), point R is at (5, 9), and point S is

sergey [27]

Answer:

(D)The midpoint of both diagonals is (4 and one-half, 5 and one-half), the slope of RP is 7, and the slope of SQ is Negative one-sevenths.

Step-by-step explanation:

Point P is at (4, 2),
Point Q is at (8, 5),
Point R is at (5, 9), and
Point S is at (1, 6)

Midpoint of SQ $=\frac{1}{2}(1+8,5+6)=(4.5,5.5)$

Midpoint of PR $=\frac{1}{2}(4+5,2+9)=(4.5,5.5)$

Now, we have established that the midpoints (point of bisection) are at the same point.

Two lines are perpendicular if the slope of one is the negative reciprocal of the other.

In option D

Slope of RP =7

Slope of SQ $=-\dfrac17$

Therefore, lines RP and SQ are perpendicular.

Option D is the correct option.

6 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

If 40% of a number is 32, what is 25% of that number?

Mekhanik [1.2K]

Well what I would do is split the 40% into 20% so from 40% to 20% that is /2. So 32/2=16 so 20% of a number is 16, we know there is 5, 20% in 100% so multiply 16 and 5 which gives you 80. Now 25% of 80 is the same as 80*.25 or 80/4 which is 20

Your Answer 20

7 0

2 years ago

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