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3 years ago

They keep returning and saying show work again

Mathematics

1 answer:

ankoles [38]3 years ago

7 0

Answer:

We conclude that:

$20\times \:3\frac{1}{5}=64$

Step-by-step explanation:

Given the expression

$20\:\times 3\frac{1}{5}$

convert mixed number to improper fraction: $a\frac{b}{c}=\frac{a\times c+b}{c}$

i.e. $3\frac{1}{5}=\frac{3\times 5+1}{5}=\frac{16}{5}$

so the expression becomes

$20\:\times 3\frac{1}{5}\:=20\times \:\frac{16}{5}$

Factor the number: 20 = 5×4

$=5\times \:4\times \frac{16}{5}$

Cross-cancel common factor: 5

$=4\times \:16$

Multiply the numbers: 4×16 = 64

$=64$

Therefore, we conclude that:

$20\times \:3\frac{1}{5}=64$

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A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than ex

Olenka [21]

Answer:

The 90% confidence interval that estimate the true proportion of students who receive financial aid is

$0.533 < p < 0.64$

$n = 1789$

Step-by-step explanation:

Considering question a

From the question we are told that

The sample size is n = 200

The number of student that receives financial aid is $k = 118$

Generally the sample proportion is

$\^ p = \frac{k}{n}$

=> $\^ p = \frac{118}{200}$

=> $\^ p = 0.59$

From the question we are told the confidence level is 90% , hence the level of significance is

$\alpha = (100 - 90 ) \%$

=> $\alpha = 0.10$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.645 * \sqrt{\frac{0.59 (1- 0.59)}{200} }$

=> $E = 0.057$

Generally 90% confidence interval is mathematically represented as

$\^ p -E < p < \^ p +E$

=> $0.533 < p < 0.64$

Considering question b

From the question we are told that

The margin of error is E = 0.03

From the question we are told the confidence level is 99% , hence the level of significance is

$\alpha = (100 - 99 ) \%$

=> $\alpha = 0.01$

Generally from the normal distribution table the critical value of is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the sample size is mathematically represented as

$[\frac{Z_{\frac{\alpha }{2} }}{E} ]^2 * \^ p (1 - \^ p )$

=> $n = [\frac{2.58}{0.03} ]^2 * 0.59 (1 - 0.59 )$

=> $n = 1789$

8 0

3 years ago

3. Mama has her opinions about things concerning Walter. Determine which statements are TRUE/FALSE about what Mama thinks.

Gnom [1K]

Answer:

False

Step-by-step explanation:

Hope this helped

5 0

3 years ago

line segment ab has a slope 5/4 and contains point a(8,-7). what is the y-coordinate of point q(3,y) if qa is perpendicular to l

MArishka [77]

U have a slope of 5/4...a perpendicular line will have a negative reciprocal slope...which means, the slope we need is -4/5.

Now we use y = mx + b
slope(m) = -4/5
(8,-7)
-7 = -4/5(8) + b
-7 = -32/5 + b
-7 + 32/5 = b
-35/5 + 32/5 = b
-3/5 = b

equation is : y = -4/5x - 3/5.....4x + 5y = -3

4x + 5y = -3....when x = 3
4(3) + 5y = -3
12 + 5y = -3
5y = -3-12
5y = - 15
y = -15/5
y = - 3 <===

7 0

3 years ago

(a(1) = -2<br> [a(n) = a(n − 1) – 5<br> Find the 4th term in the sequence.

lesya [120]

Answer:

wait of a ?

Step-by-step explanation:

7 0

3 years ago

Janice bought 3 movie tickets and 3 sodas and the total cost was $35.25. A ticket to the movie cost $5.25 more than a soda. How

Effectus [21]

Answer: a ticket to the movie costs $8.5

Step-by-step explanation:

Let x represent the cost of a movie ticket.

Let y represent the cost of a Soda.

Janice bought 3 movie tickets and 3 sodas and the total cost was $35.25. This means that

3x + 3y = 35.25 - - - - - - - - - -1

A ticket to the movie cost $5.25 more than a soda. This means that

x = y + 5.25

Substituting x = y + 25 into equation 1, it becomes

3(y + 5.25) + 3y = 35.25

3y + 15.75 + 3y = 35.25

6y = 35.25 - 15.75 = 19.5

y = 19.5/6 = 3.25

x = y + 5.25 = 3.25 + 5.25

x = 8.5

4 0

3 years ago

They keep returning and saying show work again​

They keep returning and saying show work again