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10 months ago

2. Round the following numbers to the nearest 10 thousand:

Mathematics

1 answer:

Vlad [161]10 months ago

8 0

The values of the given numbers when it is rounded up to the nearest 10 thousands are:

990000
150000

<h3>What is rounding up in mathematics?</h3>

Rounding up can be described as the process that is been used in the mathematics which is been used in the estimation of a particular number in a context.

It should be noted that in rounding the a number up, it is required to look at the next digit at the right hand of the given figures in a case whereby the digit is less than 5,the digit can be rounded down, but in the case whereby the digit is more that 5 then it can be rounded up .

From the given values, we are given the 990,201 and 159,994 and if this were to rounded up to the nearest 10 thousand then we will start from the right hand sides and round down the values less than 5 and round up the values that is more that 5. and their values will be 990000

and 150000.

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Alika [10]

Answer:

105.3

Step-by-step explanation:

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3 years ago

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A particular employee arrives at work sometime between 8:00 a.m. and 8:30 a.m. Based on past experience the company has determin

Gala2k [10]

Answer:

0.3333 = 33.33% probability that the employee will arrive between 8:15 a.m. and 8:25 a.m.

Step-by-step explanation:

A distribution is called uniform if each outcome has the same probability of happening.

The uniform distributon has two bounds, a and b, and the probability of finding a value between c and d is given by:

$P(c \leq X \leq d) = \frac{d - c}{b - a}$

A particular employee arrives at work sometime between 8:00 a.m. and 8:30 a.m.

We can consider 8 am = 0, and 8:30 am = 30, so $a = 0, b = 30$

Find the probability that the employee will arrive between 8:15 a.m. and 8:25 a.m.

Between 15 and 25, so:

$P(15 \leq X \leq 25) = \frac{25 - 15}{30 - 0} = 0.3333$

0.3333 = 33.33% probability that the employee will arrive between 8:15 a.m. and 8:25 a.m.

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2 years ago

Sadiq measured the dimensions of a 1 kg bag of sugar and found the length was 9.2 cm, the width was 6 cm and the height was 14.1

garik1379 [7]

Answer:

$d=1.28\times 10^{-6}\ g/cm^3$

Step-by-step explanation:

The mass of a bag of sugar, m = 1 kg = 0.001 g

Length of bag, l = 9.2 cm

Width of bag, b = 6 cm

Height of the bag, h = 14.1 cm

We need to find the density of the sugar bag.

Density = mass/volume

So,

$d=\dfrac{0.001\ g}{9.2\times 6\times 14.1\ cm^3}\\\\d=1.28\times 10^{-6}\ g/cm^3$

So, the density of the bag is $1.28\times 10^{-6}\ g/cm^3$ .

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2 years ago

Roy has bought two square photo frames. The ratio of their sides is 4:1. If PQ = 12 inches, what is AB? A. 12 in. B. 9 in. C. 6

aleksklad [387]

Answer:

D. 3

Step-by-step explanation:

If the ratio of their sides is 4:1, and if PQ=12 inches, then Ab=12/4=3 inches.

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3 years ago

A repeated-measures study comparing two treatments with n = 4 participants produces md = 2 and ss = 75 for the difference scores

valentinak56 [21]

The estimated standard error for the sample mean difference is 2.5 .

According to the question

A repeated-measures study comparing two treatments

n = 4

MD(mean difference) = 2

SS (sum of square) = 75

Now,

error for the sample

Formula for standard error

$S^{2} = \frac{SS}{n-1}$

by substituting the value

$S^{2} = \frac{75}{4-1}$

$S^{2} = \frac{75}{3}$

S² = 25

S = 5 (s is never negative)

Standard error of the estimate for the sample mean difference

The standard error of the estimate is the estimation of the accuracy of any predictions.

The formula for standard error of the mean difference

standard error of the mean difference = $\frac{standard\\\ error}{\sqrt{n} }$

standard error of the mean difference = $\frac{5}{\sqrt{4} }$

standard error of the mean difference = 2.5

Hence, the estimated standard error for the sample mean difference is 2.5 .

To know more about estimated standard error here:

brainly.com/question/14524236

#SPJ4

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1 year ago