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3 years ago

14

Determinati numarul natural de doua cifre pentru care suma dintre numar si rasturnatul sau e patrat perffect, iar diferenta dint

re numar si rasturnatul sau este cub perfect

1 answer:

nasty-shy [4]3 years ago

8 0

Please can you write in english

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Ky wants to buy three shirts, which were originally priced at $49.96 each. How much will the three shirts cost before tax?

tia_tia [17]

Prior to tax, each shirt added up would equal $149.88

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3 years ago

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A zoo with a total area of 982.31 acres is home to a variety of animals. If 964.36 acres have been

marishachu [46]

Answer:17.95

Step-by-step explanation:

The answer is 17.95 because you subtract 964.36 from 982.31.

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3 years ago

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A high percentage of people who fracture or dislocate a bone see a doctor for that condition. Suppose the percentage is 99%. Con

marishachu [46]

Answer:

(i) 0.15708

(ii) 0.432488

(iii) 3

Step-by-step explanation:

Given that, 99% of people who fracture or dislocate a bone see a doctor for that condition.

There is only two chance either the person having fracture or dislocation of bone will either see the doctor or not.

As per previous data, if one person got a fracture or dislocation of bone, the chance of seeing the doctor is 0.99. Assuming this chance is the same for every individual, so the total number of people having fractured or dislocated a bone can be considered as Bernoulli's population.

Let p be the probability of success represented by the chances of not seeing a doctor by any one individual having fractured or dislocated a bone.

So, p=1-0.99=0.01

According to Bernoulli's theorem, the probability of exactly r success among the total of n randomly selected from Bernoulli's population is

$P(r)=\binom{n}{r}p^r(1-p)^{n-r}\cdots(i)$

(i) The total number of persons randomly selected, n=400.

The probability that exactly 5 of them did not see a doctor

So, r=5 , p=0.01

Using equation (i),

$P(r=5)=\binom{400}{5}(0.01)^5(1-0.01)^{400-5}$

$=\frac{400!}{(400-5)!\times 5!}(0.01)^5(0.99)^{395}$

=0.15708

(ii) The probability that fewer than four of them did not see a doctor

$=P(r$

$=P(r=0)+P(r=1)+P(r=2)+P(r=3)$

$=\binom{400}{0}(0.01)^0(0.99)^{400}+\binom{400}{1}(0.01)^1(0.99)^{399}+\binom{400}{2}(0.01)^2(0.99)^{398}+\binom{400}{3}(0.01)^3(0.99)^{397}$

$=0.017951+0.072527+0.146154+0.195856$

$=0.432488$

(iii) The expected number of people who would not see a doctor

$=np$

$=300\times 0.01$

=3

7 0

3 years ago

What is the greatest common factor (gcf) of the numerator and denominator in the rational expression below? 7x+14/2x^2+x-6

Mashutka [201]

The answer is C. The numerator 7x+14 can be factored as 7(x+2). The denominator 2x^2+2x-6 can be factored as (x+2)(2x-3). The greatest common divisor of 7(x+2) and (x+2)(2x-3) is obviously x+2, which is contained in both of the two expressions.

4 0

4 years ago

In a completely randomized experimental design, three brands of paper towels were tested for their ability to absorb water. Equa

arlik [135]

Answer:

Yes. At this significance level, there is evidence to support the claim that there is a difference in the ability of the brands to absorb water.

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>The significance level is 0.05.</em>

<em>The data is:</em>

<em>Brand X: 91, 100, 88, 89</em>

<em>Brand Y: 99, 96, 94, 99</em>

<em>Brand Z: 83, 88, 89, 76</em>

We have to check if there is a significant difference between the absorbency rating of each brand.

Null hypothesis: all means are equal

$H_0:\mu_x=\mu_y=\mu_z$

Alternative hypothesis: the means are not equal

$H_a: \mu_x\neq\mu_y\neq\mu_z$

We have to apply a one-way ANOVA

We start by calculating the standard deviation for each brand:

$s_x^2=30,\,\,s_y^2=6,\,\,s_z^2=35.33$

Then, we calculate the mean standard error (MSE):

$MSE=(\sum s_i^2)/a=(30+6+35.33)/3=71.33/3=23.78$

Now, we calculate the mean square between (MSB), but we previously have to know the sample means and the mean of the sample means:

$M_x=92,\,\,M_y=97,\,\,M_z=84\\\\M=(92+97+84)/3=91$

The MSB is then:

$s^2=\dfrac{\sum(M_i-M)^2}{N-1}\\\\\\s^2=\dfrac{(92-91)^2+(97-91)^2+(84-91)^2}{3-1}\\\\\\s^2=\dfrac{1+36+49}{2}=\dfrac{86}{2}=43\\\\\\\\MSB=ns^2=4*43=172$

Now we calculate the F statistic as:

$F=MSB/MSE=172/23.78=7.23$

The degrees of freedom of the numerator are:

$dfn=a-1=3-1=2$

The degrees of freedom of the denominator are:

$dfd=N-a=3*4-3=12-3=9$

The P-value of F=7.23, dfn=2 and dfd=9 is:

$P-value=P(F>7.23)=0.01342$

As the P-value (0.013) is smaller than the significance level (0.05), the null hypothesis is rejected.

There is evidence to support the claim that there is a difference in the ability of the brands to absorb water.

3 0

4 years ago