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hjlf
3 years ago
6

Plzz help I will give brainiest

Mathematics
1 answer:
tatuchka [14]3 years ago
6 0

Answer: 99+121

220cm^2

Step-by-step explanation:

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10⁶

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Ashleyhas$40inasavingsaccount.Theinterestrateis 10% peryearandisnotcompounded.Howmuchwillshehavein4years?
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Which is which
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Answer and Explanation:

To find : Which number is in standard notation or scientific notation?

Solution :

Scientific notation is a special ways of writing the standard form of number.

Scientific notation make a big number into smaller way bye writing it into 10 to the power or using E.

Standard notation is like 12,47950585,89000.

Scientific notation is like 1.43\times 10^3 , 1.43E10

So, Now we examine the numbers,

A) 5.9E10 is the Scientific notation.

B) 0.0000734 is the Standard notation.

C) 2.66\times 10^{-3} is the Scientific notation.

D) 641,000,000,000 is the Standard notation.

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Calculate the side lengths using the given scale.
coldgirl [10]

Answer:

x = 9 m

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Step-by-step explanation:

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7 0
3 years ago
The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp
motikmotik

Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305

The degrees of freedom for this sample size are:

df=n-1=50-1=49

The t-value for a 95% confidence interval and 49 degrees of freedom is t=2.01.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.01 \cdot 0.305=0.61

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 6.34-0.61=5.73\\\\UL=M+t \cdot s_M = 6.34+0.61=6.95

The 95% confidence interval for the mean is (5.73, 6.95).

4 0
3 years ago
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