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3 years ago

How do you solve my problem

Mathematics

2 answers:

antiseptic1488 [7]3 years ago

3 0

Where is your problem

victus00 [196]3 years ago

3 0

You have to work from the inside first, then you can solve your problems.

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Robert is a high school basketball player. In a particular game, he made some three point shots and some free throws (worth one

Oksanka [162]

Answer:

let t = number of three-point shots

let f = number of free throw shots

Step-by-step explanation:

t + f = 9 (he made a total of 9 shots altogether)

3t + 1f = 23 (number of shots multiplied by their point value)

substitution method: t = 9 - f

3(9 - f) + f = 23

27 - 3f + f = 23

-2f = -4

f = 2

find 't': t + 2 = 9

t = 7

8 0

3 years ago

A deck is in the shape of a right triangle. The area of the deck is 44 m². The length of one side of the deck next to the right

Andrews [41]

11m
The area of a triangle is bh/2. If we know the height (8m), we can plug it into a=bh/2 for a=8h/2 or a=4h or a/4=h
Since we know the area (44m^2), we can put that in for 44/4=h or h=11 metres.

8 0

3 years ago

12. If the domain of f(x) = 2x+3 is (-3 << SO),

defon

2 0 owiwwkwkwkwkwkwkwkwkkwkwkw

7 0

3 years ago

Brian ate 2/3 of a pizza. His friend, Joseph, ate 1/4 of what was left. How much pizza did Joseph eat?

Natasha_Volkova [10]

Answer:

1/12

Steps:

1 - 2/3 = 1/3

1/3 / 4 = 1/3 × 1/4 = 1/12

3 0

3 years ago

A lock on a lab door has buttons numbered 0 through 9. The right code for opening the lock is 8542. Also, if on three consecutiv

Viktor [21]

Answer:

Step-by-step explanation:

given that a lock on a lab door has buttons numbered 0 through 9. The right code for opening the lock is 8542

Also, if on three consecutive attempts, a wrong code (a code is a combination of 4 ordered distinct digits) is entered, then a burglar alarm is let off.

when you forget the code, you try at random

There are 10 ways of selecting each digit

So total number of ways of trials at random = $10^4 =10000$

The probability that you get in (without letting off the alarm).

= Probability that you try correct number in I or II or III trial atmost

Prob for correct guessing = 1/10000

The probability that you get in (without letting off the alarm).

= Probability that you try correct number in I or II or III trial atmost

= $\frac{1}{10000} +\frac{9999}{10000}\frac{1}{10000}+(\frac{9999}{10000})^2\frac{1}{10000}$

Because if you get in I trial you leave. If you fail in I you try for second, similarly if you fail 2 times 3rd should be correct

=0.0002

7 0

3 years ago

How do you solve my problem​

How do you solve my problem