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Alecsey
1 month ago
13

If there are 2.2 pounds in 1 kilogram how many pounds are there in x kilograms?

Mathematics
1 answer:
tester [12.3K]1 month ago
5 0

Answer:

P=x/2.2

Step-by-step explanation:

We can set up an equation to solve this.

Let P represent the number of pounds

Let X represent the number of kilograms

Knowing that 1 kilogram equals 2.2 pounds.

So, if there is 1 kilogram (X) it translates to 2.2 pounds (P).

X=2.2P

Simply rearrange the equation to determine the pounds in x kilograms. Thus, divide both sides by 2.2.

Therefore, we derive P=x/2.2

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A certain pen has been designed so that true average writing lifetime under controlled conditions (involving the use of awriting
tester [12383]

Answer:

a) Null hypothesis:\mu \geq 10

Alternative hypothesis:\mu < 10

b) p_v =P(t_{17}

Given that the p-value is lower than the significance threshold in this situation, we have sufficient grounds to reject the null hypothesis.

c) p_v =P(t_{17}

In this case, since the p-value exceeds the significance threshold, we have adequate evidence to FAIL to reject the null hypothesis.

d) p_v =P(t_{17}

Here again, with the p-value being less than the significance level, we can reject the null hypothesis.

Step-by-step explanation:

1) Provided data and references

\bar X represents the average of the samples

s denotes the standard deviation of the samples

n=18 indicates the number of samples

\mu_o =10 is the value we are examining

\alpha defines the significance level for the test.

t represents the specific statistic of interest

p_v indicates the p-value relevant to the test (the variable of concern)

Define the null and alternative hypotheses.

To assess if the true mean is at least 10 hours, we must set up a hypothesis:

Part a

Null hypothesis:\mu \geq 10

Alternative hypothesis:\mu < 10

If we consider the sample size being less than 30 and the population deviation unknown, it’s more appropriate to use a t-test to compare the actual mean with the reference value, calculated as:

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)

Part b

In this scenario t=-2.3, \alpha=0.05

Initially, we need to calculate the degrees of freedom df=n-1=18-1=17

Since this is a left-tailed test, the p-value is determined by:

p_v =P(t_{17}

In this instance, with the p-value being less than the significance level, we have sufficient evidence to reject the null hypothesis.

Part c

For this situation t=-1.8, \alpha=0.01

We need to find the degrees of freedom df=n-1=18-1=17

For the left-tailed test, the p-value is given by:

p_v =P(t_{17}

In this case, since the p-value is above the significance level, we have enough grounds to FAIL to reject the null hypothesis.

Part d

For this case t=-3.6, \alpha=0.05

Firstly, we find the degrees of freedom df=n-1=18-1=17

Since we are conducting a left-tailed test, the p-value is calculated as:

p_v =P(t_{17}

Here, with the p-value being lower than the significance threshold, we can reject the null hypothesis.

5 0
3 months ago
Mel slides down waterslide A, and Victor slides down waterslide B. After 2 seconds, Mel was 50 feet in the air, and after 5 seco
Leona [12618]
Mel descends 50-35=15 feet in 5-2=3 seconds. Thus, her descent rate is 15/3=5 ft/sec.
Victor descends 60-50=10 feet in 4-1=3 seconds. Hence, his descent rate is 10/3=3⅓ ft/sec. Mel is faster, traveling at 5 ft/sec.
7 0
2 months ago
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Alejandro made an error in the steps below when determining the equation of the line that is perpendicular to the line 4x – 3y =
Svet_ta [12734]
The equation of the perpendicular line can be identified by determining its slope and applying the given point within the standard formula.

Standard equation: y-y1 = m(x-x1)

m*m'=-1
where m' indicates the slope of the perpendicular line
m denotes the slope of the original line

m = -coefficient of x/coefficient of y = -4/-3 = 4/3
m' = -3/4

Substituting the point (3, -2):
y+2 = -3/4*(x-3)
4y+8 = -3x+9

Thus, the equation of the perpendicular line is: 3x+4y-1=0
7 0
2 months ago
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