Ryan completed 1/8 of his test in 2/5 hour. If Ryan’s rate stayed the same, how much of his test was finished in one hour

Mathematics

1 answer:

IceJOKER [234]3 years ago

6 0

Answer:

5/16

Step-by-step explanation:

In 5/2 the time, we expect that 5/2 × 1/8 = 5/16 of the test will be complete.

_____

5/2 × (2/5 h) = 1 h

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A student's pay of $18 an hour is $4 more than twice the amount the student earns per hour at an internship. Enter and solve an

Natasha2012 [34]

1. You have that a student's pay of $18 an hour is $4 more than twice the amount the student earns per hour at an internship.

2. So, let's call the hourly pay of the internship: "x". Then, you have:

18=2x+4

3. When you clear the variable "x", you obtain:

18-4=2x
14=2x
x=14/2
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Therefore, the answer is: The hourly pay of the internship is $7.

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8(b+1)=-3(2+b) simply

gogolik [260]

Step-by-step explanation:

8(b+1)=-3(2+b)

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$1, 000 after 2 years at an interest rate of 3.0% is $ round up to the nearest cent

BaLLatris [955]

This may be a simple interest computation.

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Principal = 1,000
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term = 2 years

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3 years ago

Flow meters are installed in urban sewer systems to measure the flows through the pipes. In dry weatherconditions (no rain) the

ziro4ka [17]

Answer:

a) $\frac{(8)(30.23)^2}{20.09} \leq \sigma^2 \leq \frac{(8)(30.23)^2}{1.65}$

$363.90 \leq \sigma^2 \leq 4430.80$

Now we just take square root on both sides of the interval and we got:

$19.08 \leq \sigma \leq 66.56$

b) For this case we are 98% confidence that the true deviation for the population of interest is between 19.08 and 66.56

Step-by-step explanation:

423.6, 487.3, 453.2, 402.9, 483.0, 477.7, 442.3, 418.4, 459.0

Part a

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case we need to find the sample standard deviation with the following formula:

$s=sqrt{\frac{\sum_{i=1}^8 (x_i -\bar x)^2}{n-1}}
And in order to find the sample mean we just need to use this formula:
[tex]\bar x =\frac{\sum_{i=1}^n x_i}{n}$