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2 years ago

17.a. Three numbers have HCF of 12 and LCM 720. If two of the

Mathematics

1 answer:

ivolga24 [154]2 years ago

3 0

Answer:

Yes it is true and 4 mark not ture

Step-by-step explanation:

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Why is 1+(-5) equal to -4

erica [24]

1 + (-5) becomes 1 - 5 because you are adding a negative number.

1 - 5 is -4

5 0

3 years ago

The sum of 26 and twice a number is -46 find the number

charle [14.2K]

Answer:

-72

Step-by-step explanation:

Sum is adding so I just did substitution for this. 26+-72=-46 because a positive plus a negative equals a negative.

4 0

3 years ago

Ralph is 3 times as old as Sarah. in six years, Ralph will be only twice as old is Sarah will be then. find Ralphs age now.

Phantasy [73]

First, we start with the equation that the problem told us, which is:

R + 6 = 2 * (x + 6)

Then, we distribute the two:

R + 6 = 2x + 12

Now, we put R on its own side.

R = 2x +6

So, the answer must be C, 2x + 6

8 0

3 years ago

Read 2 more answers

Simply the expression given below (3x^3+4x^2+4x-12)+(-5x^3+9x^2-9x+20)

Svet_ta [14]

Answer:

1. Simplify: (4x2 - 2x) - (-5x2 - 8x).

Solution:

(4x2 - 2x) - (-5x2 - 8x)

= 4x2 - 2x + 5x2 + 8x.

= 4x2 + 5x2 - 2x + 8x.

= 9x2 + 6x.

= 3x(3x + 2).

Answer: 3x(3x + 2)

Step-by-step explanation:

6 0

3 years ago

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist

Tcecarenko [31]

Answer:

a. 0.7291

b. 0.9968

c. 0.7259

Step-by-step explanation:

a. np and n(1-p) can be calculated as:

$np=23\times 0.48\\\\=11.04\\\\n(1-p)=23(1-0.52)\\\\=11.96$

#Both np and np(1-p) are greater than 5, hence, normal approximation is most appropriate:

$\mu_x=11.04\\\\\sigma^2=np(1-p)=0.48\times 0.52\times 23=5.7408$

#Define Y:

Y~(11.04,5.7408)

$P(X\leq 12)\approx P(Y\leq 12.5)\\\\P(Z\leq \frac{(12.5-11.04)}{\sqrt{5.7408}})=\\\\=1-0.2709\\\\=0.7291$

Hence, the probability of 12 or fewer is 0.8291

b. The probability that 5 or more fish were caught.

#Using normal approximation:

$P(X \geq 5) \approx P(Y \geq 4.5) = P(Z \geq\frac{ (4.5-11.04)}{\sqrt{(5.7408)}})\\\\=1-0.0032\\\\=0.9968$

Hence, the probability of catching 5+ is 0.9968

c. The probability of between 5 and 12 is calculated as;

-From b above $P(X\geq 5)=0.9968$ and a , $P(X\leq 12)$ =0.7291

$P(5\leq X\leq 120\approx P(4.5\leq Y\leq 12.5)\\\\=0.7291-(1-0.9968)\\\\=0.7259$

Hence, the probability of between 5 and 12 is 0.7259

4 0

3 years ago