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2 years ago

Round off 17.2 to one significant figure

1 answer:

5 0

The answer is 17
becoz whenever the digit after the decimal is less than 5 then it is rounded off to the same number before the decimal
But when the decimal digit is more than 5 then the number is increased by 1

You might be interested in

Find the GCF of monomials 34s^2t and 96st^2

andrey2020 [161]

Answer:

Down below

Step-by-step explanation:

$3264s^{3} t^{3}$

6 0

2 years ago

Weight of a rock: In a geology course, students are learning to use a balance scale to accurately weigh rocks. One student plans

statuscvo [17]

Answer:

The student who weighted the rock 5 times has a 95% confidence interval of (25.2, 29.1) which is guaranteed to be more wider (less precise) than the other student who weighted the rock 20 times.

Step-by-step explanation:

What is Confidence Interval?

The confidence interval represents an interval that we can guarantee that the target variable will be within this interval for a given confidence level.

The confidence interval is given by

$CI = \bar{x} + t_{\alpha/2}(\frac{\sigma}{\sqrt{n} } ) \\$

Where $\bar{x}$ is the mean weight $\sigma$ is the standard deviation $t_{\alpha/2}$ is the critical value from t-table and n is the sample size.

The term $t_{\alpha/2}(\frac{\sigma}{\sqrt{n} } )$ is known as margin of error.

As the sample size is decreased the corresponding margin of error increases which results in wider confidence interval which means smaller precision.

The student who weighted the rock 5 times has a 95% confidence interval of (25.2, 29.1) which is guaranteed to be more wider (less precise) than the other student who weighted the rock 20 times.

We can say with 95% confidence that the true mean weight of the rock is within the interval of (25.2, 29.1).

7 0

2 years ago

Mikela claims that the sum of any two numbers is greater than the larger of the two numbers. Jim claims that the sum of any two

qaws [65]

<u>ANSWER: </u>

The statement of Jim is correct

<u>SOLUTION: </u>

Given statements are

Mikela claims that sum of any two numbers is greater than the larger of the two numbers.

Jim claims that the sum of any two natural numbers is greater than the larger of the two numbers.

Let a,b be the two numbers and a is the greatest of a and b

According to the mikela statement = a + b > a

b > a – a

b > 0

Here, we got an condition that b > 0, but it may not be true always.

Above condition fails when b is either 0 or negative number.

Jim - sum of any two natural numbers is greater than the larger of the two numbers.

Let a,b be the two natural numbers. And a is the greatest of a and b

According to the jim statement = a + b > a

b > a – a

b > 0

Here, we got an condition that b > 0,

And this statement will always be true because b is an natural number, so it will be greater than 0.

Hence jim’s statement is true.

8 0

2 years ago

Which table contains only corresponding x-values and y-values where the value of y is one more than the product of x and 2?

Dmitry [639]

Answer:

Option (C)

Step-by-step explanation:

Value of y is more than the product of x and 2.

Equation representing the given condition will be,

y = 2x + 1

Therefore, by substituting the values of x in the equation we can get the table for the input - output values for the equation,

x 0 1 2 3 4

y 1 3 5 7 9

Therefore, table given in the Option (C) will be the correct option.

4 0

2 years ago

A lake contains 4 distinct types of fish. Suppose that each fish caught is equally likely to be any one of these types. Let Y de

strojnjashka [21]

Answer:

a) P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90

a= μ-3.16*σ , b= μ+3.16*σ

b) P(Y≥ μ+3*σ ) ≥ 0.90

b= μ+3*σ

Step-by-step explanation:

from Chebyshev's inequality for Y

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

where

Y = the number of fish that need be caught to obtain at least one of each type

μ = expected value of Y

σ = standard deviation of Y

P(| Y - μ|≤ k*σ ) = probability that Y is within k standard deviations from the mean

k= parameter

thus for

P(| Y - μ|≤ k*σ ) ≥ 1-1/k²

P{a≤Y≤b} ≥ 0.90 → 1-1/k² = 0.90 → k = 3.16

then

P(μ-k*σ≤ Y ≤ μ+k*σ ) ≥ 0.90

using one-sided Chebyshev inequality (Cantelli's inequality)

P(Y- μ≥ λ) ≥ 1- σ²/(σ²+λ²)

P{Y≥b} ≥ 0.90 → 1- σ²/(σ²+λ²)= 1- 1/(1+(λ/σ)²)=0.90 → 3= λ/σ → λ= 3*σ

then for

P(Y≥ μ+3*σ ) ≥ 0.90

5 0

2 years ago