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3 years ago

What is 6 3/5 - 4 3/5

Mathematics

2 answers:

atroni [7]3 years ago

8 0

The answer is 2 because 6 3/5 - 4 3/5 = 2

jekas [21]3 years ago

6 0

You do 3/5 minus 3/5 which cancels out to 0. Then you do 6 minus 4 which is 2. The answer is 2.

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Tres veces un número x, restado de 18 es menor que -90

andrey2020 [161]

The answer is linked below

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2 years ago

There are 64 singers in the choir.The tenors and sopranos are in separate rows.There are 8 singers in each row.There are 4 rows

stealth61 [152]

4 rows because 4 times 8 equals 32, so then there are only 32 singers left. 64-32=32, and once again 4 times 8 equals 32.

7 0

2 years ago

Read 2 more answers

Production of a gas flow meter takes place in two distinct operations. Measurements (n = 15) of the time required for the first

Karolina [17]

Answer:

a) P(a > 80) = 0.323

b) The 95% confidence interval = (73.40, 80.60)

c) The 95% confidence interval expresses that the mean of the distribution can always be found in the given range, with a 95% confidence level.

Step-by-step explanation:

X ~ (45, 4)

Y ~ (32, 2.5)

(X+Y) ~ (77, 6.5)

Let a = (X+Y)

a) Probability that the time required to complete both of those steps will exceed 80 min = P(a > 80)

This is a normal distribution problem

We then standardize 80 min time

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (a - μ)/σ = (80 - 77)/6.5 = 0.46

To determine the probability the time required to complete both of those steps will exceed 80 min = P(a > 80) = P(z > 0.46)

We'll use data from the normal probability table for these probabilities

P(a > 80) = P(z > 0.46) = 1 - P(z ≤ 0.46) = 1 - 0.677 = 0.323

b) 95% confidence interval for the expected total time required to produce one flow meter.

We need to obtain the margin of error

Margin of error = (critical value) × (standard error of the sample)

Critical value for a 95% confidence interval = critical value for a significance level of 5% = t(15-1, 0.05/2) = 2.145 (using the t-score since information on the population mean and standard deviation isn't known)

Standard error for the sample of sum of times = (standard deviation of the sum of times)/√n = (6.5/√15) = 1.678

Margin of error = 2.145 × 1.678 = 3.60

Limits of the confidence interval = (Sample mean ± margin of error)

Lower limit of the confidence interval = (Sample mean - margin of error) = 77 - 3.60 = 73.40

Upper limit of the confidence interval = 77 + 3.60 = 80.60

The confidence interval = (73.40, 80.60)

4 0

3 years ago

Please answer this correctly without making mistakes

satela [25.4K]

Answer:

8940 table spoons and 2 teaspoons

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

Suppose that the sample standard deviation was s = 5.1. Compute a 98% confidence interval for μ, the mean time spent volunteerin

NISA [10]

Answer:

The 95% confidence interval would be given by (5.139;5.861)

Step-by-step explanation:

1) Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

2) Confidence interval

Assuming that $\bar X =5.5$ and the ranfom sample n=1086.

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=1086-1=1085$

Since the Confidence is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.01,1085)".And we see that $t_{\alpha/2}=2.33$

Now we have everything in order to replace into formula (1):

$5.5-2.33\frac{5.1}{\sqrt{1086}}=5.139$

$5.5+2.333\frac{5.1}{\sqrt{1086}}=5.861$

So on this case the 95% confidence interval would be given by (5.139;5.861) We are 98% confident that the mean time spent volunteering for the population of parents of school-aged children is between these two values.

5 0

3 years ago