2 years ago

Find two positive consecutive integers such that the square of the first added to three times the second is 24

Mathematics

1 answer:

Naddika [18.5K]2 years ago

7 0

Answer:

3 and 5

Step-by-step explanation:

Let the two consecutive odd integers be x and x + 2.

Now,

The given statement is

$x^2 + 3 \times (x+2)=24\\x^2 + 3x+6-24 = 0 \\x^2 + 3x-18 = 0\\x^2+6x-3x-18=0\\x(x+6)-3(x+6)=0\\(x-3)(x+6)=0\\x=3,-6$

Taking x=3.

Hence, two consecutive odd integers are 3 and 5.

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Use the given values of n= 93 and p= 0.24 to find the minimum value that is not significantly low, μ- 2σ , and the maximum val

allsm [11]

Answer:

The answer is "Option a".

Step-by-step explanation:

$n= 93 \\\\p= 0.24\\\\\mu=?\\\\ \sigma=?\\\\$

Using the binomial distribution: $\mu = n\times p = 93 \times 0.24 = 22.32\\\\\sigma = \sqrt{n \times p \times (1-p)}=\sqrt{93 \times 0.24 \times (1-0.24)}=4.1186$

In this the maximum value which is significantly low, $\mu-2\sigma$ , and the minimum value which is significantly high, $\mu+2\sigma$ , that is equal to: