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wel
4 years ago
12

Mary paints wooden signs and sells them at craft fairs. The function s(t) approximates how many signs Mary paints per hour. The

function w(h) represents how many hours per week Mary spends painting the wooden signs. What are the units of measurement for the composite function s(w(h))? Signs over weeks hours over weeks weeks over signs signs over hours
Mathematics
2 answers:
Furkat [3]4 years ago
7 0

Answer:

A

Step-by-step explanation:

Vinil7 [7]4 years ago
4 0

the units of measurement for the composite function s(w(h)) is  \frac{Signs}{weeks} . Correct option A) .

<u>Step-by-step explanation:</u>

Here we have , Mary paints wooden signs and sells them at craft fairs. The function s(t) approximates how many signs Mary paints per hour. The function w(h) represents how many hours per week Mary spends painting the wooden signs. We need to find What are the units of measurement for the composite function s(w(h)) . Let's find out:

Basically we have two functions as ,

s(t) : how many signs Mary paints per hour

⇒ \frac{Signs}{hour}

w(h) : how many hours per week Mary spends painting the wooden signs

⇒ \frac{hours}{week}

In function composition, if we have two function f(x) and g(x) then

(f.g)(x) or f(g(x)) means first apply g(), then apply f() i.e. applying function f to the results of function g. So , In s(w(h)) :

⇒ \frac{Signs}{hour}(\frac{hours}{week})

⇒ \frac{Signs}{weeks}

Therefore ,  the units of measurement for the composite function s(w(h)) is  \frac{Signs}{weeks} . Correct option A) .

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A coin is tossed 5 times. Find the probability that all are heads. Find the probability that at most 2 are heads.
fenix001 [56]

Answer:

1/32

15/32

Step-by-step explanation:

For a fair sided coin,

Probability of heads, P(H) = 1/2

Probability of tails P(T)  = 1/2

For a coin tossed 5 times,

P( All heads)

= P(HHHHH),

= P (H) x P(H) x P(H) x P(H) x P(H)

= (1/2) x (1/2) x (1/2) x (1/2) x (1/2)

= 1/32 (Ans)

For part B, it is easier to just list the possible outcomes for

"at most 2 heads" aka "could be 1 head" or "could be 2 heads"

"One Head" Outcomes:

P(HTTTT), P(THTTT) P(TTHTT), P(TTTHT), P(TTTTH)

"2 Heads" Outcomes:

P(HHTTT), P(HTHTT), P(HTTHT), P(HTTTH), P(THHTT), P(THTHT), P(THTTH), P(TTHHT), P(TTHTH), P(TTTHH)

If we count all the possible outcomes, we get 15 possible outcomes representing "at most 2 heads)

we know that each outcome has a probability of 1/32

hence 15 outcomes for "at most 2 heads" have a probability of

(1/32) x 15  = 15/32

8 0
3 years ago
Read 2 more answers
Help me with this math problem please
mrs_skeptik [129]

Answer:

y=8

RS=77, ST=31

Step-by-step explanation:

Note that the entire segment RT is the combined lengths of RS and ST. So:

RT=RS+ST

We know that RT is 108, RS is (9y+5), and ST is (3y+7). Substitute:

108=(9y+5)+(3y+7)

On the right, combine like terms:

108=(9y+3y)+(5+7)

Add:

108=12y+12

Subtract 12 from both sides. The right cancels:

96=12y

Divide both sides by 12:

y=8

So, the value of y is 8.

To find RS and ST, substitute 8 for y. So:

RS=9y+5

Substitute 8 for y:

RS=9(8)+5

Multiply:

RS=72+5

Add:

RS=77

Do the same for ST:

ST=3y+7

Substitute 8 for y:

ST=3(8)+7

Multiply:

ST=24+7

Add:

ST=31

5 0
4 years ago
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qaws [65]

Answer:

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VMariaS [17]
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