Solve the following equation for x: 6(4x+50)=3(x+8)+3 round to the nearest hundredth

Mathematics

1 answer:

nalin [4]3 years ago

8 0

Answer:

x = -13

Step-by-step explanation:

6(4x + 50) = 3(x+8) + 3

24x + 300 = 3x + 24 + 3

21x + 300 = 27

21x = -273

x = -13

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According to an almanac, 60% of adult smokers started smoking before turning 18 years old. (a) compute the mean and standard

larisa [96]

The given problem describes a binomial distribution with p = 60% = 0.6. Given that there are 400 trials, i.e. n = 400.

a.) The mean is given by:

$\mu=np=400\times0.6=240$

The standard deviation is given by:

$\sigma=\sqrt{npq} \\ \\ =\sqrt{np(1-p)} \\ \\ =\sqrt{400(0.6)(0.4)} \\ \\ =\sqrt{96}=9.80$

b.) The mean means that in an experiment of 400 adult smokers, we expect on the average to get about 240 smokers who started smoking before turning 18 years.

c.) It would be unusual to observe 340 smokers who started smoking before turning 18 years old in a random sample of 400 adult smokers because 340 is far greater than the mean of the distribution.
340 is greater than 3 standard deviations from the mean of the distribution.

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3 years ago

A national grocery store chain wants to test the difference in the average weight of turkeys sold in Detroit and the average wei

Arte-miy333 [17]

Answer:

Calculated value t = 1.3622 < 2.081 at 0.05 level of significance with 42 degrees of freedom

The null hypothesis is accepted .

Assume the population variances are approximately the same

Step-by-step explanation:

Explanation:-

Given data a random sample of 20 turkeys sold at the chain's stores in Detroit yielded a sample mean of 17.53 pounds, with a sample standard deviation of 3.2 pounds

The first sample size 'n₁'= 20

mean of the first sample 'x₁⁻'= 17.53 pounds

standard deviation of first sample S₁ = 3.2 pounds

Given data a random sample of 24 turkeys sold at the chain's stores in Charlotte yielded a sample mean of 14.89 pounds, with a sample standard deviation of 2.7 pounds

The second sample size n₂ = 24

mean of the second sample "x₂⁻"= 14.89 pounds

standard deviation of second sample S₂ = 2.7 pounds

Null hypothesis:-H₀: The Population Variance are approximately same

Alternatively hypothesis: H₁:The Population Variance are approximately same

Level of significance ∝ =0.05

Degrees of freedom ν = n₁ +n₂ -2 =20+24-2 = 42

Test statistic :-

$t = \frac{x^{-} _{1} - x_{2} }{\sqrt{S^2(\frac{1}{n_{1} } }+\frac{1}{n_{2} } }$