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2 years ago

Rhombus LMNO is shown with its diagonals. The length of LN is 28 centimeters. What is the length of LP? 7 cm 9 cm 14 cm 21 cm

Mathematics

2 answers:

vampirchik [111]2 years ago

7 0

LP= 1/2 * 28 = 14 Answer C.

Marrrta [24]2 years ago

7 0

Answer: The length of LP is 14 cm.

Step-by-step explanation:

Here, LMNO is a rhombus,

In which LN and MO are the diagonals, intersecting at P.

Since, the diagonals of a rhombus bisect each other,

In rhombus LMNO,

LP = PN and MP = PO,

Here, LN = 28 cm

⇒ LP + PN = 28 cm

⇒ LP + LP = 28 cm

⇒ 2 LP = 28 cm

⇒ LP = 14 cm

Thus, the length of LP is 14 cm.

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No, because she could receive as low as 14% of the vote.

No, because she could receive as low as 49% of the vote.

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Yes, because the poll stated that she will win with 63% of the vote.

Answer:

No, because she could receive as low as 49% of the vote.

Step-by-step explanation:

Given that:

Poll suggestion = 63%

Margin of Error (E) = 14%

This means that; the range or interval in which the vote obtained could fall would be ;

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63% ± 14%

Lower bound attainable = (63% - 14%) = 49%

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Since, winning actually requires obtaining atleast half of the votes (that is 50%) ; and suggested poll suggested votes could fall as low as 49% ; then Janice can't be so confident of victory.

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Step-by-step explanation:

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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 = $Z_(_\frac{\alpha}{2}_) \times \sqrt{\frac{\hat p(1-\hat p)}{n} }$