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3 years ago

15 ÷ 3 = 5. Explain why it is true that 1.5 ÷ 0.3 and 0.15 ÷ 0.03 have the same quotient.

Mathematics

1 answer:

vladimir1956 [14]3 years ago

3 0

Answer:

Explained

Step-by-step explanation:

here is a rule if the numerator and the denominator both are divided or multiplied by the same constant there no change in the quotient of the division. 15 divided by 3 will give 5 as quotient so does 1.5÷0.3 and 0.15÷0.03 because in the later both numerator and the denominator are divided by 10 and 100 and the quotient remains unchanged.

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Please help.......................

Westkost [7]

Answer:

FD = 8

Step-by-step explanation:

Since the triangles are similar then the ratios of corresponding sides are equal, that is

$\frac{FD}{BA}$ = $\frac{FE}{BC}$ , substitute values

$\frac{x+3}{2x+2}$ = $\frac{16}{24}$ = $\frac{2}{3}$ ( cross- multiply )

2(2x + 2) = 3(x + 3) ← distribute parenthesis on both sides

4x + 4 = 3x + 9 ( subtract 3x from both sides )

x + 4 = 9 ( subtract 4 from both sides )

x = 5

Thus

FD = x + 3 = 5 + 3 = 8

5 0

3 years ago

The distribution of the number of people in line at a grocery store has a mean of 3 and a variance of 9. A sample of the numbers

Airida [17]

Answer:

a) $P(\bar X >4)=P(Z>\frac{4-3}{\frac{3}{\sqrt{50}}}=2.357)$

$P(Z>2.357)=1-P(Z$

b) $P(\bar X \frac{2.5-3}{\frac{3}{\sqrt{50}}}=-1.179)$

$P(Z$

$P(2.5 < \bar X< 3.5) = P(\frac{2.5-3}{\frac{3}{\sqrt{50}}}$

$P(-1.179$ Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the number of people of a population, and for this case we know that:

Where $\mu=3$ and $\sigma=\sqrt{9}=3$

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by: