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gulaghasi [49]
3 years ago
7

The weather was a youth soccer field must be at least 45 m but it cannot exceed 60 m right to inequalities that describe the wid

th,w, of a youth soccer field. Then write two integers that are solutions of both inequalities
Mathematics
1 answer:
ss7ja [257]3 years ago
7 0

Answer:

The width of the two soccer field must be given by 45\leq w\leq 60.

Assuming two Integers that are solutions of both inequalities are 45 and 60.

Step-by-step explanation:

Let the width of the youth soccer field be 'w'.

Now Given:

the width of the youth soccer field must be at least 45 m.

So we can say that;

w\geq 45 \ m

Also Given:

the width of the youth soccer field must not exceed 60 m.

So we can say that;

x\leq 60

Hence the width of the two soccer field must be given by 45\leq w\leq 60.

So we can say that;

Any Integer lying between integers 45 and 60 can be considered as the width of the youth soccer field.

So let us Assume two Integers that are solutions of both inequalities are 45 and 60.

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Please help me answer for 35 and 36
dexar [7]

Answer:

35. f(2)=16

36. f(2)=77

Step-by-step explanation:

Question 35:

Given:

f(1)=32

f(n)=f(n-1)\times \frac{1}{2}

Plug in 2 for n and simplify. This gives,

f(2)=f(2-1)\times \frac{1}{2}\\f(2)=f(1)\times \frac{1}{2}\\f(2)=32\times \frac{1}{2}\\f(2)=16

Therefore, term 2 is 16.

Question 36:

Given:

f(1)=11

f(n)=f(n-1)\times 7

Plug in 2 for n and simplify. This gives,

f(2)=f(2-1)\times 7\\f(2)=f(1)\times 7\\f(2)=11\times 7\\f(2)=77

Therefore, term 2 is 77.

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3 years ago
Solve the value X ASAP!!
anzhelika [568]
The answer is B I think
8 0
3 years ago
The length of a rectangle is 2 cm more than four times the width. If the perimeter of the rectangle is 84 cm, what are its dimen
katovenus [111]

Answer:

Length 34 cm

Width 8 cm

Step-by-step explanation:

Let

x-----> the length of the rectangle

y----> the width of the rectangle

we know that

The perimeter of rectangle is equal to

P=2(x+y)

P=84 cm

so

84=2(x+y)

42=x+y ----> equation A

x=4y+2 -----> equation B

substitute equation B in equation A and solve for y

42=(4y+2)+y

5y=42-2

y=40/5=8 cm

Find the value of x

x=4y+2 ----> x=4(8)+2=34 cm

The dimensions are

Length 34 cm

Width 8 cm

8 0
3 years ago
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300

Step-by-step explanation:

8 0
2 years ago
Read 2 more answers
A stationary store has decided to accept a large shipment of ball-point pens if an inspection of 17 randomly selected pens yield
Leno4ka [110]

Answer:

0.762 = 76.2% probability that this shipment is accepted

Step-by-step explanation:

For each pen, there are only two possible outcomes. Either it is defective, or it is not. The probability of a pen being defective is independent from other pens. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

17 randomly selected pens

This means that n = 17

(a) Find the probability that this shipment is accepted if 10% of the total shipment is defective. (Use 3 decimal places.)

This is P(X \leq 2) when p = 0.1. So

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{17,0}.(0.1)^{0}.(0.9)^{17} = 0.167

P(X = 1) = C_{17,1}.(0.1)^{1}.(0.9)^{16} = 0.315

P(X = 2) = C_{17,2}.(0.1)^{2}.(0.9)^{15} = 0.280

P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.167 + 0.315 + 0.280 = 0.762

0.762 = 76.2% probability that this shipment is accepted

8 0
3 years ago
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