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Bezzdna [24]
3 years ago
6

The product of two consecutive negative integers is 600. What is the value of the lesser integer?

Mathematics
2 answers:
lapo4ka [179]3 years ago
8 0
X ·( x + 1 ) = 600
x² + x = 600
x² + x - 600 = 0
x_{12} = \frac{-1+/- \sqrt{1+2400} }{2}= \frac{-1+/- \sqrt{2401} }{2}= \\  =\frac{-1+/-50}{2}
x = ( - 1 - 49 ) / 2 = - 50 : 2 = - 25
x = - 25,  x + 1 = - 24
( - 25 ) · ( - 24 ) = 600
Answer:
C )   - 25
Likurg_2 [28]3 years ago
5 0

x ·( x + 1 ) = 600

x² + x = 600

x² + x - 600 = 0

x_{12} = \frac{-1+/- \sqrt{1+2400} }{2}= \frac{-1+/- \sqrt{2401} }{2}= \\  =\frac{-1+/-50}{2}  

x = ( - 1 - 49 ) / 2 = - 50 : 2 = - 25

x = - 25,  x + 1 = - 24

( - 25 ) · ( - 24 ) = 600

Answer:

C )   - 25


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(a) Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the prop
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Answer:

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(b) The sample sizes are 6666.

Step-by-step explanation:

(a)

The information provided is:

Confidence level = 98%

MOE = 0.02

n₁ = n₂ = n

\hat p_{1} = \hat p_{2} = \hat p = 0.50\ (\text{Assume})

Compute the sample sizes as follows:

MOE=z_{\alpha/2}\times\sqrt{\frac{2\times\hat p(1-\hat p)}{n}

       n=\frac{2\times\hat p(1-\hat p)\times (z_{\alpha/2})^{2}}{MOE^{2}}

          =\frac{2\times0.50(1-0.50)\times (2.33)^{2}}{0.02^{2}}\\\\=6786.125\\\\\approx 6787

Thus, the sample sizes are 6787.

(b)

Now it is provided that:

\hat p_{1}=0.45\\\hat p_{2}=0.58

Compute the sample size as follows:

MOE=z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})}{n}

       n=\frac{(z_{\alpha/2})^{2}\times [\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})]}{MOE^{2}}

          =\frac{2.33^{2}\times [0.45(1-0.45)+0.58(1-0.58)]}{0.02^{2}}\\\\=6665.331975\\\\\approx 6666

Thus, the sample sizes are 6666.

7 0
3 years ago
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