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2 years ago

True or False? A circle could be circumscribed about the quadrilateral below.

Mathematics

2 answers:

Aleks04 [339]2 years ago

4 0

Answer:

The answer is False.

Step-by-step explanation:

Given a quadrilateral ABCD with angles ∠A=90°, ∠B=90°, ∠C=115° and ∠D=65°

We have to tell about the statement true or false

'A circle could be circumscribed about the quadrilateral'

As, we say a circle can be circumscribed about the quadrilateral if and only if the quadrilateral is cyclic and also if the opposite angles of a quadrilateral are supplementary, then only the quadrilateral is cyclic.

Hence, we have to check for the opposite angles of quadrilateral if they are supplementary then we say the above statement is true i.e A circle could be circumscribed about the quadrilateral.

∠A+∠C=90+115=205° ≠ 180°

⇒ opposite angles are not supplementary.

∴ A circle could not be circumscribed about the quadrilateral.

The above statement is false.

BlackZzzverrR [31]2 years ago

3 0

Answer: false

Step-by-step explanation: the answer is false because suplimentary angled dont add up to 180

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EleoNora [17]

Answer:

18 years

Step-by-step explanation:

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$A = P(1 + \frac{r}{n})^{nt}$

Note that r is percentage converted to decimal. So 3% = 3/100 = 0.03

We can rearrange the above equation to:

$\frac{A}{P} = (1 + \frac{r}{n})^{nt}$

Taking logs on both sides

$log(\frac{A}{P}) = log(1 + \frac{r}{n})^{nt}$

This gives

$log(\frac{A}{P}) =nt \times log(1 + \frac{r}{n})\\So,\\nt = \frac{log(\frac{A}{P})}{ log(1 + \frac{r}{n})}$

In this particular problem, n = 4, , A= 9600, P = 5600, r =0.03, so r/n = 0.03/4 = 0.0075

1 + r/n = 1+0.0075 = 1.0075

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2 years ago

A box contains 13 red blocks, 8 orange blocks, 9 green blocks, and 10 blue blocks. Anna randomly chooses a block from the box.

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9. An express delivery service advertises next-day delivery of any parcel it receives before 3PM. The actual success rate is 90%

Juli2301 [7.4K]

Answer:

a) P(X =16 ) = 0.1853

b) $P(X \leq 12)$ = 0.0684

Step-by-step explanation:

GIVEN DATA:

n = 16

p = 0.90

from relation given probabllity can be solve

$P(X) = ^nC_x * p^x * ( 1 - p)^{n-x}$

$P(X =16 ) = ^{16}C_{16} * 0.90^x * ( 1 - 0.90)^{16-16}$

P(X =16 ) = 0.1853

b) $P(X \leq 12) = 1 - P(X \geq 13)$

= 1 - [ P(X = 13) +P(X = 14) +P(X = 15) +P(X = 16) ]

$= 1 - [ ^{16}C_{13} * 0.90^{13} * (1 - 0.90)^3 +^{16}C_{14} * 0.90^{14} * (1 - 0.90)^2 +^{16}C_{15} * 0.90^{15} * (1 - 0.90)^1 +^{16}C_{16} * 0.90^{16} * (1 - 0.90)^0 ]$

= 0.0684

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