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3 years ago

Three decimals between 8.6 and 9.2?

Mathematics

2 answers:

charle [14.2K]3 years ago

6 0

Answer:

8.7, 8.9, 9.0

Step-by-step explanation:

nydimaria [60]3 years ago

4 0

Answer:

8.7, 8.9, & 9.1

Step-by-step explanation:

The decimal places go up by tens like all other numbers. Therefore, 8.7, 8.9, and 9.1 are all in between 8.6 and 9.2

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Mashcka [7]

Answer:

Step-by-step explanation:

6 times 3 equals 18, then you subtract 3 from that and thats the answer

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2 years ago

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Given f(x) = x^3 - 2x^2 - x + 2, the roots of f(x) are

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3 years ago

Suppose that A and B are square matrices and that ABC is invertible. Show that each of A, B, and C is invertible.

Murljashka [212]

Answer:

You can proceed as follows:

Step-by-step explanation:

Suppose that the matrix $ABC$ is invertible, and suppose that at least one of the matrices $A,B,C$ is not invertible. Without loss of generality suppose that the matrix $A$ is not invertible. Remember the important result that a matrix is invertible if and only if its determinant is nonzero. Then,

$\det (ABC)\neq 0.$

On the other hand, the determinant of a products of matrices is the product of the determinants of the matrices, that is to say,

$\det (ABC)=\det(A)\cdot \det(B)\cdot \det (C).$

But we supposed that $A$ is not invertible. Then $\det (A)=0$ . Then $\det(A)\cdot \det(B)\cdot \det (C)=0$ . This contradicts the fact that

$\det (ABC)=\det(A)\cdot \det(B)\cdot \det (C)$

and then the three matrices $A,B,\, \text{and}\, C$ must be invertible matrices.

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3 years ago

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Anuta_ua [19.1K]

Answer:

The answer is c

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4 years ago

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The weights of a certain dog breed are approximately normally distributed with a mean of 49 pounds, and a standard deviation of

AlekseyPX

Answer:

a. 74.86%

b. 50%

c. 50%

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with a mean of 49 pounds, and a standard deviation of 6 pounds.

This means that $\mu = 49, \sigma = 6$

a. Find the percentage of dogs of this breed that weigh less than 53 pounds.

The proportion is the p-value of Z when X = 53. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{53 - 49}{6}$

$Z = 0.67$

$Z = 0.67$ has a p-value of 0.7486.

0.7486*100% = 74.86%, which is percentage of dogs of this breed that weigh less than 53 pounds.

b. Find the percentage of dogs of this breed that weigh less than 49 pounds.

p-value of Z when X = 49, so:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{49 - 49}{6}$

$Z = 0$

$Z = 0$ has a p-value of 0.5

0.5 = 50% of dogs of this breed that weigh less than 49 pounds.

c. Find the percentage of dogs of this breed that weigh more than 49 pounds.

1 subtracted by the p-value of Z when X = 49, so:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{49 - 49}{6}$

$Z = 0$

$Z = 0$ has a p-value of 0.5.

1 - 0.5 = 0.5 = 50% of dogs of this breed that weigh more than 49 pounds.

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3 years ago