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3 years ago

Two times a number plus 5 equals 25. What is the number?

2 answers:

8 0

This is actually very simple to figure out. What times 2 equals 20? The answer is 10. 10 times 2 equals 20. Add the 5, and you have 25.
Your answer is 10

Svetlanka [38]3 years ago

3 0

The number is ten. hope it helps

You might be interested in

The owner of an orange grove estimates that if 24 trees are planted per acre, then each mature tree will yield 600 oranges per y

tia_tia [17]

Answer:

16428 oranges

Explanation:

Total yield = number of trees × number of oranges in each tree

Initial yield = 600×24= 14400 oranges

To find the equation needed, let x = additional trees and y= total yield

Number of trees = 24 +x

Number of oranges in each tree = 600-12x

Equation of total yield y= (24+x)(600-12x)

y= 14400-288x+600x-12x²

y= -12x²+312x+14400

Using a graphing calculator, from the graph drawn for this quadratic equation, we notice that it is a parabola. Therefore to find the maximum value, we should find the maximum point which is at the vertex of the parabola, we use the formula x= -b/2a

A quadratic equation is such: ax²+bx+c

Therefore x =-312/2×-12

x= -312/-24

x= 13

So we can conclude that in order to maximise oranges from the trees, the person needs to plant an additional 13 trees. Substituting from the above:

24+x=24+13= 37 trees in total

y= -12x²+312x+14400= -12×13²+312×13+14400= -2028+4056+14400

=16428 oranges in total yield

8 0

3 years ago

PLZZZ HELP DUE IN 5 MINS

iris [78.8K]

Answer:

{(2, 9), (5, 9), (1, -6), (-5, 5), (2,9), (5,9), (1,−6), (−5,5)}

Step-by-step explanation:

recall that for a function to be valid, every input must give one and only 1 output. (i.e every x-ordinate must have only one unique corresponding y-ordinate)

I have highlighted in bold and underlined the entries for each of the other choices where an x-value gives more than one possible y-value, hence these are not functions. only the first option satisfies this condition:

{(2, 9), (5, 9), (1, -6), (-5, 5), (2,9), (5,9), (1,−6), (−5,5)} Every x-value has a unique y-value, so it is possible that this is a function.

{(6, 6), (6, 8), (-8, -7), (-3, -8), (6,6),(6,8),(−8,−7),(−3,−8)} (not a function)

{(1, 8), (9, -3), (7, -4), (7, -6), (1,8),(9,−3),(7,−4),(7,−6)} (not a function)

{(-5, -1), (-3, -7), (-5, 3), (-7, -4), (−5,−1),(−3,−7),(−5,3),(−7,−4)} (not a function)

4 0

3 years ago

A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d

tankabanditka [31]

The selection of r objects out of n is done in

$C(n, r)= \frac{n!}{r!(n-r)!}$ many ways.

The total number of selections 10 that we can make from 6+7=13 students is

$C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}= 286$
thus, the sample space of the experiment is 286

A.
"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.

 $C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35$

P(all 6 girls chosen)=35/286=0.12

B.
"The probability that a randomly chosen team has 3 girls and 7 boys."

with the same logic as in A, the number of groups were all 7 boys are in, is

 $C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20$

so the probability is 20/286=0.07

C.
"The probability that a randomly chosen team has either 4 or 6 boys."

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is 0.12.

case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls.

 $C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105$

the probability is 105/286=0.367

since case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"The probability that a randomly chosen team has 5 girls and 5 boys."

selecting 5 boys and 5 girls can be done in

 $C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126$

many ways,

so the probability is 126/286=0.44

6 0

3 years ago

Read 2 more answers

Need help............

svp [43]

Answer:

i would say C

Step-by-step explanation:

6 0

3 years ago

7+3 to the second power+(12-8) divided by 2x 4 is

cupoosta [38]

$\bf \stackrel{\mathbb{P~E~M~D~A~S}}{7+3^2+(12-8)\div 2\times 4}\implies 7+3^2+(\stackrel{\downarrow }{4})\div 2\times 4\implies 7+\stackrel{\downarrow }{9}+(4)\div 2\times 4 \\\\\\ 7+9+\stackrel{\downarrow }{2}\times 4\implies 7+9+\stackrel{\downarrow }{8}\implies \stackrel{\downarrow }{16}+8\implies 24$

5 0

3 years ago