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2 years ago

Which of the following statements is false?

Mathematics

2 answers:

kenny6666 [7]2 years ago

6 0

Answer:

A). True

B). False

C). True

D). True

Step-by-step explanation:

A).Quadrilateral means any polygon with four sides or four edges. Hence a square is a regular quadrilateral with four equal sides or edges.

B). Given any quadrilateral ABCD, BC and CD are adjacent and not BC and AD.

C). A rectangle is an equiangular quadrilateral. Equiangular means having equal angles. Rectangle have all the angles equal of 90 degree. Other example of equiangular quadrilateral is a square.

D). By theorem of parallelogram, if a quadrilateral is a parallelogram, then it has two sets of opposite sides congruent.

nikklg [1K]2 years ago

4 0

Answer:

B. Given quadrilateral ABCD, BC⎯⎯⎯⎯⎯⎯⎯⎯ and AD⎯⎯⎯⎯⎯⎯⎯⎯ are adjacent

Step-by-step explanation:

A. A square has 4 congruent sides and 4 congruent angles, so it is a regular polygon. True.

B. Draw the quadrilateral and label the vertices in the order shown. Given quadrilateral ABCD, BC and AD are opposite sides, not adjacent sides. False.

C. All angles of a rectangle are right angles that measure 90 degrees. True.

D. True by a theorem.

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A baby otter was born 3/4 of a month early. At birth it's weight was 7/8 kilograms which is 9/10 kilogram less than the average

Assoli18 [71]

Answer:

The average weight of new born otter was, $1\frac{31}{40}$

Step-by-step explanation:

Let average weight of new born otter be x.

As per the given statement: At birth it's weight was 7/8 kilograms which is 9/10 kilogram less than the average weight of a new born otter in the aquarium

" $\frac{9}{10}$ kg less than average weight of a new born otter" means $x-\frac{9}{10}$

As per the given information, we have;

$\frac{7}{8} = x -\frac{9}{10}$

Add $\frac{9}{10}$ both sides, we have;

$\frac{7}{8} + \frac{9}{10} = x$

Take LCM of 8 and 10 is, 40

⇒ $\frac{35+36}{40}=x$

Simplify:

$x = \frac{71}{40} = 1\frac{31}{40}$

Therefore, the average weight of new born otter was, $1\frac{31}{40}$

6 0

2 years ago

Read 2 more answers

Helena uses a mixture of compost and topsoil in her garden. She purchased a total of 10 cubic yards of compost and topsoil for $

Annette [7]

Let C be the amount of compost

T be the amount of topsoil

Each compost cost = $25

Cost of C compost = 25C

Each topsoil cost = $15

Cost of T topsoil = 15T

Amount of compost + amount of topsoil = 10

C + T = 10 -------> Equation 1

cost of C compost + cost of T topsoil = 180

25C + 15T = 180 --------> equation 2

Solve the first equation for C

C + T = 10

C = 10 - T

Now plug it in second equation

25C + 15T = 180

25 ( 10 - T) +15T = 180

250 - 25T + 15T = 180 (combine like terms)

250 - 10 T = 180 (Subtract 250 on both sides)

-10T = 180 - 250

-10T = -70 ( divide by -10 on both sides)

T = 7

She purchased 7 cubic yards of topsoil .

4 0

2 years ago

Read 2 more answers

The CEO of a local tech company wants to estimate the proportion of employees that are pleased with the cleanliness of the break

Zigmanuir [339]

Answer:

Methods of obtaining a sample of 600 employees from the 4,700 workforce:

Part A: The type of sampling method proposed by the CEO is Convenience Sampling.

Part B: When there are equal number of participants in both campuses, stratification by campus would give a more precise approximation of the proportion of employees who are satisfied with the cleanliness of the breakrooms than stratification by gender. Another method to ensure that stratification by campus gives a more precise approximation of the proportion of employees who are satisfied with the cleanliness of the breakrooms than stratification by gender is to ensure that the sample is proportional to the proportion of each campus to the whole population or workforce.

Step-by-step explanation:

A Convenience Sampling technique is a non-probability (non-random) sampling method and the participants are selected based on availability (early attendees). The early attendees might be different from the late attendees in characteristics such as age, sex, etc. Therefore, sampling biases are present. All non-probability sampling methods are prone to volunteer bias.

Stratified sampling is more accurate and representative of the population. It reduces sampling bias. The difficulty arises in choosing the characteristic to stratify by.

3 0

2 years ago

Mike and Jasmine have a combined age of 30. Mike is 3 years younger than twice Jasmines age. How old is Mike?

Kamila [148]

Mike is 19.

If you turn the question into an equation it turns to:
J x 2 - 3 = M

So since you don’t know Jasmine’s age you would use guess and check. 8,9, and 10 didn’t work but when you put in 11:
11 x 2 - 3 = M
22 - 3 = M
19 = M

And to check your work
J + M has to equal 30 so,
11 + 19 = 30
30 = 30

6 0

2 years ago

John's commute to work is 20kmhr while Sheri's commute is 500mmin. Who has the fastest commute to work in mihrif 1.61km=1mi? A S

Andreyy89

Answer:

A) Sheri has the faster commute by 6.2 miles/hr.

Step-by-step explanation:

Given

John's commute to work $=20\ km/hr$

Sheri's commute to work $=500\ m/min$

$1.61\ km = 1\ mile$

John's commute to work in miles per hour = $\frac{20\ km}{1 hr}\times \frac{1\ mile}{1.61\ km}= 12.42\ miles/hr$

Sheri's commute to work in miles per hour = $\frac{500\ m}{1\ min}\times \frac{1\ km}{1000\ m}\times \frac{1\ mile}{1.61\ km}\times \frac{60\ min}{1\ hr}= 18.63\ miles/ hr$

We can see that Sheri has a faster commute.

Difference between the rates = $18.63\ miles/ hr-12.42\ miles/hr=6.21\ miles/hr\approx 6.2\ miles/hr$

∴ Sheri has the faster commute by 6.2 miles/hr.

3 0

2 years ago