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2 years ago

Can someone help me plz

Mathematics

1 answer:

galben [10]2 years ago

6 0

Step-by-step explanation:

the answer is in the image above

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Josiah went to the local barber to get his hair cut. It cost $18 for the haircut. Josiah tipped the barber 15%. What was the tot

Inessa [10]

The total cost of the haircut including the tip is $20.70. 18*15%=2.70+18=20.7.

Hope this helps:)

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3 years ago

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I need the answer asap .! help please !

shtirl [24]

Answer:

i tried the math and if im correct it should be 0.52N

Step-by-step explanation:

im sorry if its not correct bro

8 0

2 years ago

Can someone please help me asap !!!!!!.???

Tatiana [17]

A, C, D, F? Not sure but took a good guess

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3 years ago

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Lotteries are an important income source for various governments around the world. However, the availability of lotteries and ot

SashulF [63]

Answer: 0.0473

Step-by-step explanation:

Given : The proportion of gambling addicts : p=0.30

Let x be the binomial variable that represents the number of persons are gambling addicts.

with parameter p=0.30 , n= 10

Using Binomial formula ,

$P(x)=^nC_xp^x(1-p)^{n-x}$

The required probability = $P(x>5)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\=^{10}C_6(0.3)^6(0.7)^{4}+^{10}C_7(0.3)^7(0.7)^{3}+^{10}C_8(0.3)^8(0.7)^{2}+^{10}C_9(0.3)^9(0.7)^{1}+^{10}C_{10}(0.3)^{10}(0.7)^{0}\\\\=\dfrac{10!}{4!6!}(0.3)^6(0.7)^{4}+\dfrac{10!}{3!7!}(0.3)^7(0.7)^{3}+\dfrac{10!}{2!8!}(0.3)^8(0.7)^{2}+\dfrac{10!}{9!1!}(0.3)^9(0.7)^{1}+(1)(0.3)^{10}\\\\=0.0473489874\approx0.0473$

Hence, the required probability = 0.0473

5 0

2 years ago

A veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses

Licemer1 [7]

Answer:

Probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

Step-by-step explanation:

We are given that a veterinary researcher takes a random sample of 60 horses presenting with colic. The average age of the random sample of horses with colic is 12 years. The average age of all horses seen at the veterinary clinic was determined to be 10 years. The researcher also determined that the standard deviation of all horses coming to the veterinary clinic is 8 years.

So, firstly according to Central limit theorem the z score probability distribution for sample means is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = average age of the random sample of horses with colic = 12 yrs

$\mu$ = average age of all horses seen at the veterinary clinic = 10 yrs

$\sigma$ = standard deviation of all horses coming to the veterinary clinic = 8 yrs

n = sample of horses = 60

So, probability that a sample mean is 12 or larger for a sample from the horse population is given by = P( $\bar X$ $\geq$ 12)

P( $\bar X$ $\geq$ 12) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{12-10}{\frac{8}{\sqrt{60} } }$ ) = P(Z $\geq$ 1.94) = 1 - P(Z < 1.94)

= 1 - 0.97381 = 0.0262

Therefore, probability that a sample mean is 12 or larger for a sample from the horse population is 0.0262.

4 0

3 years ago