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frosja888 [35]
3 years ago
12

What is 8.546 rounded to nearest tens place​

Mathematics
2 answers:
igomit [66]3 years ago
7 0
8.5 bc the 4 is telling the 5 to stay. If it was a 5 instead of a 4 then it would be 8.6 don’t listen to the other person I’m correct :)
zhannawk [14.2K]3 years ago
5 0

Answer:

8.6

Step-by-step explanation:

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A baseball weighs 5 ounces. How much do 6 baseballs weigh?
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This is a question that involves multiplication. If 1 baseball weighs 5 ounces, and you're trying to find out how many 6 weigh, multiply 5x6. The answer is 30. Hope this helps.
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Where did Jordan make her mistake? GIVING BRAINLIEST!
Ivenika [448]

Answer:

Jordan made mistake in his first step while multiplying.

Step-by-step explanation:

Given expression of Jordan's work is

14(8+4x) = 6x-(5-11x)

In the first step, Jordan multiplied the numbers to remove the braces.

He multiplied 14(8+4x) wrong as 14*8 is 112 rather than 126and on the right hand side, he did not change the sign of 11x to positive as two negatives are positive when multiplied.

Hence,

Jordan made mistake in his first step while multiplying.

7 0
3 years ago
A company buys computers and printers. Each computer costs $550 and each printer costs $390. If the company spends $8160 and buy
maxonik [38]

Answer:the company bought 12 computers and 4 printers.

Step-by-step explanation:

Let x represent the number of computers that the company bought.

Let y represent the number of printers that the company bought.

The company buys a total of 16 machines. It means that

x + y = 16

Each computer costs $550 and each printer costs $390. If the company spends $8160 for all the computers and printers that was bought, it means that

550x + 390y = 8160 - - - - - - - - - - 1

Substituting x = 16 - y into equation 1, it becomes

550(16 - y) + 390y = 8160

8800 - 550y + 390y = 8160

- 550y + 390y = 8160 - 8800

- 160y = - 640

y = - 640/ - 160

y = 4

Substituting y = 4 into x = 16 - y, it becomes

x = 16 - 4

x = 12

6 0
3 years ago
on the asymptotic behavior of the sample estimates of eigenvalues and eigenvectors of covariance matrices
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A complex mathematical topic, the asymptotic behavior of sequences of random variables, or the behavior of indefinitely long sequences of random variables, has significant ramifications for the statistical analysis of data from large samples.

The asymptotic behavior of the sample estimators of the eigenvalues and eigenvectors of covariance matrices is examined in this claim. This work focuses on limited sample size scenarios where the number of accessible observations is comparable in magnitude to the observation dimension rather than usual high sample-size asymptotic .

Under the presumption that both the sample size and the observation dimension go to infinity while their quotient converges to a positive value, the asymptotic behavior of the conventional sample estimates is examined using methods from random matrix theory.

Closed form asymptotic expressions of these estimators are obtained, demonstrating the inconsistency of the conventional sample estimators in these asymptotic conditions, assuming that an asymptotic eigenvalue splitting condition is satisfied.

To learn more about asymptotic behavior visit:brainly.com/question/17767511

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1 year ago
70. Suppose that you currently have one credit card with a balance of $10,000 at an annual rate of 24.00% interest. You have sto
Simora [160]

Answer:

1+r+r^{2}+r^{3}+.....+r^{n-1} = \frac{1-r^{n}}{1-r}

Step-by-step explanation:

Taking the succession:

1+r+r^{2}+r^{3}+.....+r^{n-1}

You can multiply and divide by 1-r without chaging the result:

\frac{1-r}{1-r} (1+r+r^{2}+r^{3}+.....+r^{n-1})

Distributing the upper part of the fraction you have:

\frac{1}{1-r} (1-r+r-r^{2}+r^{2}-r^{3}+r^{3}-r^{4}+.....+r^{n-1}-r^{n})

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1+r+r^{2}+r^{3}+.....+r^{n-1} = \frac{1-r^{n}}{1-r}

5 0
3 years ago
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