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3 years ago

4 8/9 divided by 2/3

Mathematics

2 answers:

Veseljchak [2.6K]3 years ago

5 0

Answer:

Step-by-step explanation:

Same denominator and then just divide.

Anuta_ua [19.1K]3 years ago

4 0

Answer:

$\frac{22}{3}$

Step-by-step explanation:

$\frac{44}{9} \times \frac{3}{2}$

$\frac{22}{9} \times 3$

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gavmur [86]

Given that 2 similar signs have a scale factor of 5:3 and the larger sign is 45 ft, then the smaller sign will have the height given by:
scale factor=(height of large sign )/(height of smaller sign)
let the height of smaller sign be x, plugging in the values we get:
5/3=45/x
solving for x we get
3/5=x/45
hence
x=3/5×45
x=27 ft
The height of smaller sign will be 27 ft

7 0

3 years ago

Does anybody know how to make a pie chart out of this

garri49 [273]

Answer:

Step-by-step explanation:

since it is 100% and the pie chart circle is 360°, then for each % it will be 3.6°

applying this into drawing, just times 3.6° to the percentage and then get the final angle for the data.

--------------------------------------------

if you don't need to be so concise, just see as you fit

3 0

4 years ago

David made 30 cups of tea per hour. Andrew made 35 cups of coffee per hour. This week David made an additional 10 cups of tea. W

olga2289 [7]

X = cups of tea per hour
y = cups of coffee per hour

David makes 30 cups of tea per hour = 30x
David works for D hours this week.

Therefore number of weekly cups of tea = number of hours * cups of tea per hour
This is 30x * D = 30Dx

For Andrew, it is similar
30Ay
Because Andrew made an additional 10 cups, then it is 30Ay + 10
(cups of coffee this week)

4 0

3 years ago

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

The sample proportion is $\r p = 0.47$

$E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

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

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

3 0

3 years ago

What of the following best describes the advantage of having data presented in a table over a paragraph

Yuki888 [10]

What are the choices?

6 0

4 years ago