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3 years ago

Please help meee I need help

Mathematics

1 answer:

natta225 [31]3 years ago

8 0

Answer:

Step-by-step explanation:

To solve, collect like terms

x ≥ 2 + 21/2

x ≥ 4 1/2

the whole number that would be greater than 41/2 would be 5

2. 2.7 + 3.3 > x

6 > x

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Paul has 144 army man and his toy box into six equal groups for battle how many army men are in each group

iogann1982 [59]

I'm pretty sure this one is super easy
you just take 144 and divide it by 6 which would look like
144/6 then the answer is 24

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3 years ago

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Help me solve this pleasee

tangare [24]

Because Angle A and B are supplementary they will ad up to 180 degrees.
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3 years ago

What are the degrees of freedom associated with the factor(s) in this study design?

miss Akunina [59]

Answer:

The question is unclear and incomplete.

Let me explain the degrees of freedom in statistics.

Step-by-step explanation:

Statistically, degrees of freedom which is denoted as DF is the number of independent values that can vary in an analysis without breaking any constraints. It can also be referred to as the number of independent values that a statistical analysis can estimate.

Degrees of freedom also define the probability distributions for the test statistics of various hypothesis tests.

The degree of freedom has the formula:

DF = N - 1 where N number of random variables

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3 years ago

Pls help me and thank youuu

nlexa [21]

Answer:

x = 1

Step-by-step explanation:

14 -4x = 10

-4x = -4

x = 1

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2 years ago

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The sodium content of a popular sports drink is listed as 200 mg in a 32-oz bottle. Analysis of 20 bottles indicates a sample me

julia-pushkina [17]

Answer:

The sample does not contradicts the manufacturer's claim.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 200 mg

Sample mean, $\bar{x}$ = 211.5 mg

Sample size, n = 20

Alpha, α = 0.05

Sample standard deviation, s = 18.5 mg

a) First, we design the null and the alternate hypothesis for a two tailed test

$H_{0}: \mu = 200\\H_A: \mu \neq 200$

We use Two-tailed t test to perform this hypothesis.

b) Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$ Putting all the values, we have

$t_{stat} = \displaystyle\frac{211.5 - 200}{\frac{18.5}{\sqrt{20}} } = 2.7799$

c) Now,

$t_{critical} \text{ at 0.05 level of significance, 19 degree of freedom } = \pm 2.0930$

Since, the calculated t-statistic does not lie in the range of the acceptance region(-2.0930,2.0930), we reject the null hypothesis.

Thus, the sample does not contradicts the manufacturer's claim.

d) P-value = 0.011934

Since the p-value is less than the significance level, we reject the null hypothesis and accept the alternate hypothesis.

Yes, both approach the critical value and the p-value approach gave the same results.

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3 years ago