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3 years ago

HOLA FELLOW AMERICANS I AWAIT YOUR ASSISTANCE

Mathematics

2 answers:

givi [52]3 years ago

7 0

Answer:

Step-by-step explanation:

sesenic [268]3 years ago

6 0

The answer is d x=3 and y=2

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What is the Range of this data

musickatia [10]

The range of this data is 65 since the minimum is 20 and the maximum is 85(if its counting by 5s) then you just subtract and you have the range.

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2 years ago

What is the perimeter of the figure? <br> 229 4/9 ft<br> 229 2/3 ft<br> 230 2/3 ft <br> 231 ft

mote1985 [20]

The answer would be 230 2/3, because to get the perimeter you have to add all the sides together, 60 5/6 + 59 1/3 + 56 1/6 + 54 1/3 = 230 2/3

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2 years ago

A prticular type of tennis racket comes in a midsize versionand an oversize version. sixty percent of all customers at acertain

svetlana [45]

Answer:

a) P(x≥6)=0.633

b) P(4≤x≤8)=0.8989 (one standard deviation from the mean).

c) P(x≤7)=0.8328

Step-by-step explanation:

a) We can model this a binomial experiment. The probability of success p is the proportion of customers that prefer the oversize version (p=0.60).

The number of trials is n=10, as they select 10 randomly customers.

We have to calculate the probability that at least 6 out of 10 prefer the oversize version.

This can be calculated using the binomial expression:

$P(x\geq6)=\sum_{k=6}^{10}P(k)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\geq6)=0.2508+0.215+0.1209+0.0403+0.006=0.633$

b) We first have to calculate the standard deviation from the mean of the binomial distribution. This is expressed as:

$\sigma=\sqrt{np(1-p)}=\sqrt{10*0.6*0.4}=\sqrt{2.4}=1.55$

The mean of this distribution is:

$\mu=np=10*0.6=6$

As this is a discrete distribution, we have to use integer values for the random variable. We will approximate both values for the bound of the interval.

$LL=\mu-\sigma=6-1.55=4.45\approx4\\\\UL=\mu+\sigma=6+1.55=7.55\approx8$

The probability of having between 4 and 8 customers choosing the oversize version is:

$P(4\leq x\leq 8)=\sum_{k=4}^8P(k)=P(4)+P(5)+P(6)+P(7)+P(8)\\\\\\P(x=4) = \binom{10}{4} p^{4}q^{6}=210*0.1296*0.0041=0.1115\\\\P(x=5) = \binom{10}{5} p^{5}q^{5}=252*0.0778*0.0102=0.2007\\\\P(x=6) = \binom{10}{6} p^{6}q^{4}=210*0.0467*0.0256=0.2508\\\\P(x=7) = \binom{10}{7} p^{7}q^{3}=120*0.028*0.064=0.215\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\\\P(4\leq x\leq 8)=0.1115+0.2007+0.2508+0.215+0.1209=0.8989$

c. The probability that all of the next ten customers who want this racket can get the version they want from current stock means that at most 7 customers pick the oversize version.

Then, we have to calculate P(x≤7). We will, for simplicity, calculate this probability substracting P(x>7) from 1.

$P(x\leq7)=1-\sum_{k=8}^{10}P(k)=1-(P(8)+P(9)+P(10))\\\\\\P(x=8) = \binom{10}{8} p^{8}q^{2}=45*0.0168*0.16=0.1209\\\\P(x=9) = \binom{10}{9} p^{9}q^{1}=10*0.0101*0.4=0.0403\\\\P(x=10) = \binom{10}{10} p^{10}q^{0}=1*0.006*1=0.006\\\\\\P(x\leq 7)=1-(0.1209+0.0403+0.006)=1-0.1672=0.8328$

7 0

2 years ago

Given: x - 5 > -2. Choose the solution set. {x | x R, x > -7} {x | x R, x > -3} {x | x R, x > 3} {x | x R, x > 7}

sweet-ann [11.9K]

Answer:

{x | x ∈ R, x > 3}

Step-by-step explanation:

We are given the following inequality:

x - 5 > -2

In order to find the solution, we have to isolate the x on one side. This can be done by adding 5 to both sides of the inequality as shown below:

x - 5 + 5 > -2 + 5

x > 3

So the solution of the inequality is set of all numbers where x > 3. This is represented by 3rd option.

So, {x | x ∈ R, x > 3} is the answer.

4 0

3 years ago

Pls help with number 4b

Leya [2.2K]

Answer:

where is the hint

Step-by-step explanation:

you might want to try reading the hint

7 0

3 years ago