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34kurt
3 years ago
15

If the area of a garden is 11 square feet what could be the dimensions of the garden

Mathematics
1 answer:
sergejj [24]3 years ago
7 0
Assuming that the garden is rectangular in shape, the area of a rectangle is given by length x width.

11 is a prime with factors 1 and 11.

Therefore, if the area of a (rectangular) garden is 11 square feet, then the possible dimension of the garden is 1 feet by 11 feet.
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Stars that are medium-sized are: A. about the size of our sun B. about four times the size of our sun C. less than half the size
aliina [53]
C because the sun is like 100 times bigger than our Earth so if its medium well there we go
8 0
3 years ago
Find the value of x. round to the nearest tenth. 9,4, x
natali 33 [55]

Answer:

Not sure what you mean here, is this a table? If so, you need to provide the y value or a screenshot!

3 0
3 years ago
Read 2 more answers
Which of the following demonstrates how the 20 is calculated using the<br> combination pattern?
ZanzabumX [31]

Answer:

D

Step-by-step explanation:

The diagram shows Pascal's triangle. Pascal's triangle is a triangular array of the binomial coefficients.

The entry in the n^{th} row (start counting rows from 0) and k^{th} column (start counting columns from 0) of Pascal's triangle is denoted by

C^n_k=\left(\begin{array}{c}n\\ k\end{array}\right)

Coefficient 20 stands in 6th row, then n = 6 and in 3rd column, so k = 3.

Hence,

20=C^6_3=\left(\begin{array}{c}6\\ 3\end{array}\right)=\dfrac{6!}{3!(6-3)!}

4 0
4 years ago
68.6943 rounded to tbe nearest tenth
lutik1710 [3]

The answer to this is 68.7

The tenths place is the first term right after the decimal point. We see a "6" there. Look the number right after that. There is a "9". Since 9 is greater than 5, we round up the "6" to "7".

3 0
3 years ago
Read 2 more answers
A simple random sample of 60 households in city 1 is taken. In the sample, there are 45 households that decorate their houses wi
murzikaleks [220]

Answer:

The calculated value of z = - 0.197  falls in the critical region therefore we reject the null hypothesis and conclude that  at  the 5% significance level there is significant difference in population proportions of households that decorate their houses with lights for the holidays

Step-by-step explanation:

We formulate the null and alternative hypotheses as

H0: p1= p2 there is no difference in population proportions of households that decorate their houses with lights for the holidays

against Ha : p1≠ p2  (claim)  ( two sided)

The significance level is set at ∝= 0.05

The critical value for two tailed test at alpha=0.05 is ± 1.96

or Z∝= 0.05/2= ± 1.96

The test statistic is

Z = p1-p2/√pq(1/n1 +1/n2)

p1= proportions of households  decorating in city 1  = 45/60=0.75

p2= proportions of households  decorating in city 2 = 40/50= 0.8

p = the common proportion on the assumption that the two proportion are same.

p =   \frac{n_1p_1 +n_2p_2}{n_1+n_2}

Calculating

p =60 (0.75) + 50 (0.8) / 110

p=  45+ 40/110= 85/110 = 0.772

so  q = 1-p= 1- 0.772= 0.227

Putting the values in the test statistic and calculating

z= 0.75- 0.8/ √0.772*0.227( 1/60 + 1/50)

z= -0.05/√ 0.175244 ( 110/300)

z= -0.05/0.25348

z= -0.197

The calculated value of z = - 0.197  falls in the critical region therefore we reject the null hypothesis and conclude that  at  the 5% significance level there is significant difference in population proportions of households that decorate their houses with lights for the holidays

4 0
3 years ago
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