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3 years ago

Does it matter what order you add the numbers in a problem? explain how chips and numbers lines support your answer.

Mathematics

1 answer:

Mrac [35]3 years ago

4 0

Answer: No, it does not matter.

Step-by-step explanation:

In addition the order does not matter, this is because of the commutative property of the addition, that says that: A + B = B + A

and the associative property, that says that:

A + (B + C) = (A + B) + C

for example, if you have 4 chips and your friend gives you her 5 chips, now you have a total of 4 + 5 = 9 chips.

The case is the same if you have 5 chips and your friend gives you 4 chips, now you have 5 + 4 = 9 chips.

Other example is that:

5 chips + ( 2chips + 1 chip) = 5 chips + 3 chips = 8 chips

(5 chips + 2 chips) + 1 chip = 7 chips + 1 chip = 8 chips

You can see that the order does not matter.

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Name the parts of the following expression: 2x+3-4x+7

lisov135 [29]

Answer:

Step-by-step explanation:

2x and -4x are like terms because they both have an x attached.3 and 7 are also like term.the number 2 is a coefficient because attached to x.-4 is also a coefficient

4 0

3 years ago

Best Answer gets brainliest, measurement isn't my cup of tea in math.

blagie [28]

The answer is 24in^2

3 0

3 years ago

Read 2 more answers

Help me solve,Brainlist for first to answer

Vanyuwa [196]

Answer:

(i) the other two sides are 6 and 6 $\sqrt{2}$

(ii) the other two sides are $\frac{4}{3} and \frac{8}{3}$

Step-by-step explanation:

(i) Sine: sin(θ) = Opposite ÷ Hypotenuse

Cosine: cos(θ) = Adjacent ÷ Hypotenuse

Tangent: tan(θ) = Opposite ÷ Adjacent

Here adjacent side = 6

opposite side = d

angle = 45°

other angles are 90° and 45°

tan (45) = Opposite ÷ Adjacent

1 = d ÷ 6

∴ d = 6 × 1 = 6

so opposite side = 6

Hypotenuse ² = opposite side ² + adjacent side²

= 6² + 6²

= 36 + 36

= 72

hypotenuse = $\sqrt{72}$

= 6 $\sqrt{2}$

the other two sides are 6 and 6 $\sqrt{2}$

(ii) here adjacent side = 4√3

angle = 30°

other angles are 90° and 60°

opposite side = d

tan ( 30) = opposite ÷ adjacent

$\frac{1}{\sqrt{3}}$ = d ÷ 4√3

$\frac{1}{\sqrt{3}}$ = d × ( $\frac{\sqrt{3}}{4}$ )

3 d = 4

therefore d = $\frac{4}{3}$

therefore opposite side = $\frac{4}{3}$

Hypotenuse ² = opposite side ² + adjacent side²

=( $\frac{4}{3}$ )² +( $\frac{4}{\sqrt{3}}$ )²

= $\frac{64}{9}$

therefore hypotenuse = $\sqrt{\frac{64}{9}}$

= $\frac{8}{3}$

the other two sides are $\frac{4}{3} and \frac{8}{3}$

7 0

4 years ago

An unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6

SIZIF [17.4K]

Answer:

Probability of rolling at least 4 sixes is 0.01696.

Step-by-step explanation:

We are given that an unbalanced die is manufactured so that there is a 20% chance of rolling a “six." The die is rolled 6 times.

The above situation can be represented through binomial distribution;

$P(X = r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,.......$

where, n = number trials (samples) taken = 6 trials

r = number of success = at least 4

p = probability of success which in our question is probability of

rolling a “six", i.e; p = 0.20

Let X = Number of sixes on a die

So, X ~ Binom(n = 6, p = 0.20)

Now, Probability of rolling at least 4 sixes is given by = P(X $\geq$ 4)

P(X $\geq$ 4) = P(X = 4) + P(X = 5) + P(X = 6)

= $\binom{6}{4} \times 0.20^{4} \times (1-0.20)^{6-4}+\binom{6}{5} \times 0.20^{5} \times (1-0.20)^{6-5}+\binom{6}{6} \times 0.20^{6} \times (1-0.20)^{6-6}$

= $15 \times 0.20^{4} \times 0.80^{2}+6 \times 0.20^{5} \times 0.80^{1}+1 \times 0.20^{6} \times 0.80^{0}$

= 0.0154 + 0.00154 + 0.000064

= 0.01696

Therefore, probability of rolling at least 4 sixes is 0.01696.

8 0

3 years ago

The male-to-female ratio in Taiwan is 52 to 48.

Neko [114]

Male to female=52:48
write as faction
52/48
divide 2
26/24
divide 2
13/12

7 0

3 years ago