All categories

1 year ago

A square garden has trees 30m,50m and 40m at three successive corners.What is the area of the land ?

Mathematics

3 answers:

Masteriza [31]1 year ago

8 0

15m because if we add all of the you will get 12 but atleast you need some for backyard

Guest1 year ago

0 0

Answer:

1500

Step-by-step explanation:

The total number of drinks times the percent of drinks that are frappuccinos equals the number of frappuccinos sold. Let d = total number of drinks sold

d *24% = 360

Change to decimal form

d * .24 = 360

Divide each side by .24

.24d/.24 = 360/.24

d = 1500

There were 1500 drinks sold

Guest

1 year ago

Guest1 year ago

0 0

Lizzie has already written 120 words of an essay. She writes 5 words per minute. Write an expression to represent the total number of words she has written after her writing session today. Let m represent the number of minutes passed during today’s session.

You might be interested in

Two processes are used to produce forgings used in an aircraft wing assembly. Of 200 forgings selected from process 1, 10 do not

tangare [24]

Answer:

$\bar p_1=0.05\\\bar p_2=0.067$

b-Check illustration below

c.(-0.0517,0.0177

Step-by-step explanation:

a.let $p_1 \& p_2$ denote processes 1 & 2.

For $p_1$ : T1=10,n1=200

For $p_2$ :T2=20,n2=300

Therefore

$\bar p_1=\frac{t_1}{N_1}=\frac{10}{200}=0.05\\\bar p_2=\frac{t_2}{N_2}=\frac{20}{300}=0.067$

b. To test for hypothesis:-

$H_0:p_1=p_2\\H_A=p_1\neq p_2\\\alpha=0.05$

ii.For a two sample Proportion test

$Z=\frac{\bar p_1-\bar p_2}{\sqrt(\bar p(1-\bar p)(\frac{1}{n_1}+\frac{1}{n_2})}\\$

iii. for $\frac{\alpha}{2}=(-1.96,+1.96)$ (0.5 alpha IS 0.025),

reject $H_o$ if $|Z|>1.96$

iv. Do not reject $H_o$ . The noncomforting proportions are not significantly different as calculated below:

$z=\frac{0.050-0.067}{\sqrt {(0.06\times0.94)\times \frac{1}{500}}}$

z=-0.78

c. $(1-\alpha).100\%$ for the p1-p2 is given as:

$(\bar p_1-\bar p_2)\pm Z_0_._5_\alpha \times \sqrt \frac{ \bar p_1(1-\bar p_1)}{n_1}+\frac{\bar p_2(1-\bar p_2)}{n_2}\\\\=(0.05-0.067)\pm 1.645 \times \sqrt \ \frac{0.05+0.95}{200}+\frac{0.067+0.933}{300}\\$

=(-0.0517,+0.0177)

*CI contains o, which implies that proportions are NOT significantly different.

4 0

2 years ago

A high school track is shaped as a rectangle with a half circle on either side.

earnstyle [38]

Answer:

Step-by-step explanation:

perimeter=2(85)+π(12.5)=220+2×12.5π

=170+25×3.14

=170+78.5

=248.5

distance ran=3×248.5=745.5 yards.

8 0

2 years ago

Read 2 more answers

Of the 125 people in the community choir, 22 of them actually come from other communities. What fraction of the choir comes from

makvit [3.9K]

The answer is 17.6%
125/100=1.25*17.6=22
hope this helps!

3 0

2 years ago

Events M and N are independent events. In this scenario, if P(M) = 0.46 and P(M and N) = 0.138, then P(N) = .

Oksi-84 [34.3K]

Answer:

P(N)=0.3

Step-by-step explanation:

Given: P(M)= 0.46, P(M and N)=0.138

Using P(M) ×P(N)= P(M and N)

⇒0.46×P(N)= 0.138

⇒P(N)= $\frac{0.138}{0.46}$

⇒P(N)=0.3

7 0

2 years ago

Read 2 more answers

Deal with these relations on the set of real numbers: R₁ = {(a, b) ∈ R² | a > b}, the "greater than" relation, R₂ = {(a, b) ∈

uranmaximum [27]

Answer:

a) R1ºR1 = R1

b) R1ºR2 = R1

c) R1ºR3 = $\{ (a,b) \in R^2 \}$

d) R1ºR4 = $\{ (a,b) \in R^2 \}$

e) R1ºR5 = R1

f) R1ºR6 = $\{ (a,b) \in R^2 \}$

g) R2ºR3 = $\{ (a,b) \in R^2 \}$

h) R3ºR3 = R3

Step-by-step explanation:

R1ºR1

(a,c) is in R1ºR1 if there exists b such that (a,b) is in R1 and (b,c) is in R1. This means that a > b, and b > c. That can only happen if a > c. Therefore R1ºR1 = R1

R1ºR2

This case is similar to the previous one. (a,c) is in R1ºR2 if there exists b such that (a,b) is in R2 and (b,c) is in R1. This means that a ≥ b, and b > c. That can only happen if a > c. Hence R1ºR2 = R1

R1ºR3

(a,c) is in R1ºR3 if there exists b such that a c. Independently of which values we use for a and c, there always exist a value of b big enough so that b is bigger than both a and c, fulfilling the conditions. We conclude that any pair of real numbers are related.

R1ºR4

This is similar to the previous one. Independently of the values (a,c) we choose, there is always going to be a value b big enough such that a ≤ b and b > c. As a result any pair of real numbers are related.

R1ºR5

If a and c are related, then there exists b such that (a,b) is in R5 and (b,c) is in R1. Because of how R5 is defined, b must be equal to a. Therefore, (a,c) is in R1. This proves that R1ºR5 = R1

R1ºR6

The relation R6 is less restrictive than the relation R3, if we find 2 numbers, one smaller than the other, in particular we find 2 different numbers. If we had 2 numbers a and c, we can find a number b big enough such that ac. In particular, b is different from a, so (a,b) is in R6 and (b,c) is in R1, which implies that (a,c) is in R1ºR6. Since we took 2 arbitrary numbers, then any pair of real numbers are related.

R2ºR3

This is similar to the case R1ºR3, only with the difference that we can take b to be equal to a as long as it is bigger than c. We conclude that any pair of real numbers are related.

R3ºR3

If a and c are real numbers such that there exist b fulfilling the relations a < b and b < c, then necessarily a < c. If a < c, then we can use any number in between as our b. Therefore R3ºR3 = R3

I hope you find this answer useful!

5 0

2 years ago

A square garden has trees 30m,50m and 40m at three successive corners.What is the area of the land ?​

A square garden has trees 30m,50m and 40m at three successive corners.What is the area of the land ?