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3 years ago

Katie started knitting at 11:27 a.m.

Mathematics

2 answers:

Bond [772]3 years ago

3 0

Answer:

2:22 pm

Step-by-step explanation:

To solve this problem, let's divide up the timing by activity so we know how long each event took:

Knitting for 2 hours and 21 minutes and starting at 11:27 am would mean Katie finished knitting at 1:48 pm.

5 minute break would end at 1:52 pm.

30 minute walk would end at 2:22 pm.

Hope this helped!

Aleksandr-060686 [28]3 years ago

3 0

Answer:

it will be 2:23pm.

Step-by-step explanation:

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Prove the following identity

Deffense [45]

Answer:

sec(x)/(tan xsin(x))=cot^2 x+1 = Ture

Step-by-step explanation:

Verify the following identity:

sec(x)/(tan(x) sin(x)) = cot(x)^2 + 1

Hint: | Eliminate the denominator on the left hand side.

Multiply both sides by sin(x) tan(x):

sec(x) = ^?sin(x) tan(x) (cot(x)^2 + 1)

Hint: | Express both sides in terms of sine and cosine.

Write cotangent as cosine/sine, secant as 1/cosine and tangent as sine/cosine:

1/cos(x) = ^?sin(x)/cos(x) sin(x) ((cos(x)/sin(x))^2 + 1)

Hint: | Simplify the right hand side.

((cos(x)/sin(x))^2 + 1) sin(x) (sin(x)/cos(x)) = (((cos(x)^2)/(sin(x)^2) + 1) sin(x)^2)/(cos(x)):

1/cos(x) = ^?(sin(x)^2 (cos(x)^2/sin(x)^2 + 1))/cos(x)

Hint: | Put the fractions in cos(x)^2/sin(x)^2 + 1 over a common denominator.

Put cos(x)^2/sin(x)^2 + 1 over the common denominator sin(x)^2: cos(x)^2/sin(x)^2 + 1 = (cos(x)^2 + sin(x)^2)/sin(x)^2:

1/cos(x) = ^?sin(x)^2/cos(x) (cos(x)^2 + sin(x)^2)/sin(x)^2

Hint: | Cancel down ((cos(x)^2 + sin(x)^2) sin(x)^2)/(sin(x)^2 cos(x)).

Cancel sin(x)^2 from the numerator and denominator. ((cos(x)^2 + sin(x)^2) sin(x)^2)/(sin(x)^2 cos(x)) = (sin(x)^2 (cos(x)^2 + sin(x)^2))/(sin(x)^2 cos(x)) = (cos(x)^2 + sin(x)^2)/cos(x):

1/cos(x) = ^?(cos(x)^2 + sin(x)^2)/cos(x)

Hint: | Eliminate the denominators on both sides.

Multiply both sides by cos(x):

1 = ^?cos(x)^2 + sin(x)^2

Hint: | Use the Pythagorean identity on cos(x)^2 + sin(x)^2.

Substitute cos(x)^2 + sin(x)^2 = 1:

1 = ^?1

Hint: | Come to a conclusion.

The left hand side and right hand side are identical:

Answer: (identity has been verified)

3 0

2 years ago

What is the area? Need help fast please

Tpy6a [65]

Greetings! Hope this helps!

Answer

A = 157

Explanation

3.14 x 10^2

3.14 x 100

314

314/2

157

Have a good day!

_______________

A brainliest would help tons! :D

6 0

3 years ago

What is 80 divided by 4

xxMikexx [17]

80 divided by 4 is 20

7 0

3 years ago

Which expression is NOT equivalent to 3^5?

Westkost [7]

Answer:

APart 1:

A.53 =53

B.3^5/3^-6 =177147

C.(1/4)^3 • (1/4)^2 = 1 /1024

D.(-7)^5/(-7)^7 = 1/49

Step-by-step explanation:

I tried my best for part 2 but i'm pretty sure it's not right. I think 1/1024 and 1/49 has a value between 0 and 1

6 0

3 years ago

A 5-card hand is dealt from a perfectly shuffled deck. Define the events: A: the hand is a four of a kind (all four cards of one

TiliK225 [7]

In a hand of 5 cards, you want 4 of them to be of the same rank, and the fifth can be any of the remaining 48 cards. So if the rank of the 4-of-a-kind is fixed, there are $\binom44\binom{48}1=48$ possible hands. To account for any choice of rank, we choose 1 of the 13 possible ranks and multiply this count by $\binom{13}1=13$ . So there are 624 possible hands containing a 4-of-a-kind. Hence A occurs with probability

$\dfrac{\binom{13}1\binom44\binom{48}1}{\binom{52}5}=\dfrac{624}{2,598,960}\approx0.00024$

There are 4 aces in the deck. If exactly 1 occurs in the hand, the remaining 4 cards can be any of the remaining 48 non-ace cards, contributing $\binom41\binom{48}4=778,320$ possible hands. Exactly 2 aces are drawn in $\binom42\binom{48}3=103,776$ hands. And so on. This gives a total of

$\displaystyle\sum_{a=1}^4\binom4a\binom{48}{5-a}=886,656$

possible hands containing at least 1 ace, and hence B occurs with probability

$\dfrac{\sum\limits_{a=1}^4\binom4a\binom{48}{5-a}}{\binom{52}5}=\dfrac{18,472}{54,145}\approx0.3412$

The product of these probability is approximately 0.000082.

A and B are independent if the probability of both events occurring simultaneously is the same as the above probability, i.e. $P(A\cap B)=P(A)P(B)$ . This happens if

the hand has 4 aces and 1 non-ace, or
the hand has a non-ace 4-of-a-kind and 1 ace

The above "sub-events" are mutually exclusive and share no overlap. There are 48 possible non-aces to choose from, so the first sub-event consists of 48 possible hands. There are 12 non-ace 4-of-a-kinds and 4 choices of ace for the fifth card, so the second sub-event has a total of 12*4 = 48 possible hands. So $A\cap B$ consists of 96 possible hands, which occurs with probability

$\dfrac{96}{\binom{52}5}\approx0.0000369$

and so the events A and B are NOT independent.

4 0

3 years ago