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TiliK225 [7]
3 years ago
15

The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives 16 years; the

Mathematics
1 answer:
STatiana [176]3 years ago
3 0

Using the Emperical rule:

68% lie with one one standard deviation:

16 + 1.7 , 16-1.7 = 17.7, 14.3

14.3 is part of the 68%.

The remaining 32% of the distribution is outside the range, with half being less than and half being greater than.

32/2 = 16

The probability of living loner than 14.3 Would be 16%

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Statement: 2x-3&gt;7 which is greater? 5 or x
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The demand for a daily newspaper at a newsstand at a busy intersection is known to be normally distributed with a mean of 150 an
AURORKA [14]

Answer:

171 newspapers.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 150, \sigma = 25

How many newspapers should the newsstand operator order to ensure that he runs short on no more than 20% of days

The number of newspapers must be on the 100-20 = 80th percentile. So this value if X when Z has a pvalue of 0.8. So X when Z = 0.84.

Z = \frac{X - \mu}{\sigma}

0.84 = \frac{X - 150}{25}

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3 years ago
Write the polynomial in factored form. <br><br> p(x)=(x+5)(x-___)(x+___)
Rom4ik [11]

Answer:


Step-by-step explanation:

Given : p(x)=x^{3} +6x^{2} -7x-60

Solution :

Part A:

First find the potential roots of p(x) using rational root theorem;

So, \text{Possible roots} =\pm\frac{\text{factors of constant term}}{\text{factors of leading coefficient}}

Since constant term = -60

Leading coefficient = 1

\text{Possible roots} =\pm\frac{\text{factors of 60}}{\text{factors of 1}}

\text{Possible roots} =\pm\frac{1,2,3,4,5,6,10,12,15,20,60}{1}

Thus the possible roots are  \pm1, \pm2, \pm 3, \pm4, \pm5,\pm6, \pm10, \pm12, \pm15, \pm20, \pm60

Thus from the given options the correct answers are -10,-5,3,15

Now For Part B we will use synthetic division

Out of the possible roots we will use the root which gives remainder 0 in synthetic division :

Since we can see in the figure With -5 we are getting 0 remainder.

Refer the attached figure

We have completed the table and have obtained the following resulting coefficients: 1 , 1,−12,0.  All the coefficients except the last one are the coefficients of the quotient, the last coefficient is the remainder.

Thus the quotient is x^{2} +x-12

And remainder is 0 .

So to get the other two factors of the given polynomial we will solve the quotient by middle term splitting

x^{2} +x-12=0

x^{2} +4x-3x-12=0

x(x+4)-3(x+4)=0

(x-3)(x+4)=0

Thus x-3 and x+4 are the other two factors

So , p(x)=(x+5)(x-3)(x+4)





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