Use a commutative property to complete the statement 2x + 16

I need help pleaseee):

ankoles [38]

Answer:

Step-by-step explanation:

7 0

2 years ago

What is the rule for translating 4 units to the right

maks197457 [2]

Answer:

When you move down you subtract and when you move to the right you subtract when your moving up or to the left you add. hope that helped :-)

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Rachel has 90p. Simon has £2.

natulia [17]

Answer:

9:20 is the answers for the question

Step-by-step explanation:

please tell me your

6 0

2 years ago

Read 2 more answers

If n+h/5=f+9/9, then n/5=

Ann [662]

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Question
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$\boxed { \frac{n+h}{5} = \frac{f+9}{9}}$

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Split the fraction on the left
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$\boxed { \frac{n}{5} + \frac{h}{5} = \frac{f + 9}{9}}$

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Take away h/5 from both sides
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$\boxed { \frac{n}{5} = \frac{f+9}{9} - \frac{h}{5}}$

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Change the denominator to be the same
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$\boxed { \frac{n}{5} = \frac{5f+45}{45} - \frac{9h}{45}}$

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Put it into single fraction
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$\boxed { \frac{n}{5} = \frac{5f+45-9h}{45} }$

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Rearrange (This step may not be necessary)
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$\boxed {\frac{n}{5} = \frac{5f-9h+ 45}{45} }$

$\bf \Longrightarrow \ Answer \ : \boxed {\boxed {\frac{n}{5} = \frac{5f-9h+ 45}{45} }}$

6 0

2 years ago

To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil

Debora [2.8K]

Answer:

statistic value t= 5.43

p-value: < 0.00001.

Step-by-step explanation:

Hello!

To obtain information over the corrosion-resistance properties of a certain type of steel conduit a random sample of 45 specimens was taken and buried for two years.

The study variable is:

X: Max. penetration of a steel conduit.

The data of the sample

n= 45

sample mean X[bar]= 53.4

sample standard deviation S= 4.2

The conduits are manufactured to have a true average penetration of at most 50 mills, symbolically: μ ≤ 50

The hypothesis is:

H₀: μ ≤ 50

H₁: μ > 50

α: 0.05

To choose the corresponding statistic to use to study the population mean, the variable must have a normal distribution. There is no available information to check this, so I'll just assume that the variable has a normal distribution and, since the population variance is unknown and the sample is small, the statistic to use is a Student t.

Under the null hypothesis, the critical region and the p-value are one-tailed.

Critical value:

$t_{n-1; 1-\alpha } = t_{44; 0.95} = 1.68$

Rejection rule:

Reject the null hypothesis when t ≥ 1.68

t= $\frac{53.4 - 50}{\frac{4.2}{\sqrt{45} } }$

t= 5.43

The calculated value is greater than the critical value, thedecision is to rject the null hypothesis.

p-value:

P(t ≥ 5.43) = 1 - P(t < 5.43) = < 0.00001.

The p-value is less than α so the decision is to reject the null hypothesis.

Since the null hypothesis was rejected, then the population average of the penetration of the conduits specimens is greater than 50 mils. It is not recommendable to use these conduits.

I hope it helps!

4 0

2 years ago