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ahrayia [7]
2 years ago
15

There are 25 trees on the Jackson’s property. Twenty percent of the trees are oak trees. Which equation can be used to find the

number of oak trees on the property?
StartFraction 25 divided by 1 Over 20 divided by 1 EndFraction = StartFraction 25 Over 20 EndFraction
StartFraction 100 times 5 Over 20 times 5 EndFraction = StartFraction 500 Over 100 EndFraction
StartFraction 20 times 4 Over 25 times 4 EndFraction = StartFraction 80 Over 100 EndFraction
StartFraction 20 divided by 4 Over 100 divided by 4 EndFraction = StartFraction 5 Over 25 EndFraction
Mathematics
1 answer:
ahrayia [7]2 years ago
8 0

Answer:

StartFraction 25 divided by 1 Over 20 divided by 1 EndFraction = StartFraction 25 Over 20 EndFraction

Step-by-step explanation:

Number of trees = 25

Percentage of aok trees = 20%

To obtain the Number of oak trees :

(Number of trees ÷ 1) ÷ (percentage ÷1)

(25 / 1) ÷ (20 / 1) = (25 /1) * (20 / 1) = 25 / 20

StartFraction 25 divided by 1 Over 20 divided by 1 EndFraction = StartFraction 25 Over 20 EndFraction

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James has established the following monthly budget for himself: Rent . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
8090 [49]

Answer:

James's monthly income is $1250.

Step-by-step explanation:

If the budget accounts for James's entire monthly income, the sum of the cost of all the items on the budget is his total monthly income.

Income = 300 + 175 + 125 + 200 + 250 + 50 + 25 + 125 = $1250

6 0
2 years ago
Read 2 more answers
Alex is drawing rectangles with different areas on a centimeter grid . He can draw 3 different rectangles with an area of 12cm s
mojhsa [17]

Answer:

Number of rectangles  could alex draw with an area of 11cm² = 1

Step-by-step explanation:

Minimum length in centimeter grid = 1 cm

Alex is drawing rectangles with different areas on a centimetre grid.He can draw 3 different rectangles with an area of 12cm²

That is

                  12 = 1 \times12\\\\12=2\times6\\\\12=3 \times4

         These are the 3 different rectangles with an area of 12cm².

Now we need to find how many rectangles could alex draw with an area of 11cm².

                    11 = 1 x 11

So only one factorization is possible.

Number of rectangles  could alex draw with an area of 11cm² = 1

6 0
1 year ago
Read 2 more answers
Annabelle measures the sides of 15 right triangles and, based on her observations, she concludes that for any right triangle the
saul85 [17]

Answer: Annabelle is using  the a measure of central tendency defined as the Mode.

Step-by-step explanation: A measure of central tendency in its simplest definition is a single value or measure that can safely be used to represent all members belonging to an entire set of given data. Hence, as a good illustration, one figure can be used to confidently represent all other ninety nine figures where a set of one hundred figures were given.

The mean, median and mode are commonly accepted measures of central tendency.

The mode is the most frequently occurring value in a given set of data. As such, the modal value is statistically acceptable as a representative of the entire set of values or data.

If Annabelle measures the sides of 15 right triangles and based on her observations, she concludes that for any right  triangle the sum of the squares of the two legs is equal to the square of the hypotenuse, what she has done is taking the most frequently occurring value, and in her experiment, the most frequent of all observed data satisfies the Pythagorean Theorem.

That is why Annabelle can confidently make her assumption.

6 0
2 years ago
Suppose we want to choose 2 objects, without replacement, from the 5 objects pencil, eraser, desk, chair, and lamp. (a)How many
BigorU [14]

Answer:

a) 20 ways

b) 10 ways

Step-by-step explanation:

When the order of selection/choice matters, we use Permutations to find the number of ways and if the order of selection/choice does not matter, we use Combinations to find the number of ways.

Part a)

We have to chose 2 objects from a group of 5 objects and order of choice matters. This is a problem of permutations, so we have to find 5P2

General formula of permutations of n objects taken r at time is:

nPr=\frac{n!}{(n-r)!}

Using the value of n=5 and r=2, we get:

5P2=\frac{5!}{(5-2)!} =20

Therefore, we can choose 2 objects from a group of 5 given objects if the order of choice matters.

Part b)

Order of choice does not matter in this case, so we will use combinations to find the number of ways of choosing 2 objects from a group of 5 objects which is represented by 5C2.

The general formula of combinations of n objects taken r at a time is:

nCr=\frac{n!}{r!(n-r)!}

Using the value of n=5 and r=2, we get:

5C2=\frac{5!}{2!(5-2)!} =10

Therefore, we can choose 2 objects from a group of 5 given objects if the order of choice does not matters.

3 0
2 years ago
Read 2 more answers
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

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

<em />

<u>So, 90% confidence interval for the population proportion, p is ;</u>

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

<u>90% confidence interval for p</u> = [ \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
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