How to rename the number 600,000

Mathematics

2 answers:

Temka [501]4 years ago

7 0

Six hundred thousand

lilavasa [31]4 years ago

4 0

Word form:
six hundred thousand

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I NEED HELP PLEASE

Elan Coil [88]

Answer:

y=x+8 because your friend's allowance is y but also $8 more than yours, x. x+8 is her allowance so x+8=y

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3 years ago

A ball is thrown off the roof of the school. Its path can be described by the equation: h(t) = -2t^2 + 8t + 10, where h is heigh

zimovet [89]

Answer:

10 meters

Step-by-step explanation:

The given function $h(t)=-2t^2+8t+10$ represents the height, $h$ , in meters, $t$ seconds after the ball is thrown.

Since the ball is thrown off the roof, then ball's height will be equal to the roof's height before being thrown (0 seconds). Therefore, substitute $t=0$ into the given function $h(t)=-2t^2+8t+10$ :

$h(0)=-2(0^2)+8(0)+10,\\h(0)=0+0+10,\\h(0)=\boxed{10\text{ meters}}$

Therefore, the building is 10 meters tall.

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3 years ago

The Dow Jones Industrial Average has had a mean gain of 432 pear year with a standard deviation of 722. A random sample of 40 ye

trasher [3.6K]

Answer:

66.98% probability that the mean gain for the sample was between 250 and 500.

Step-by-step explanation:

To solve this problem, it is important to know the Normal probability distribution and the Central limit theorem.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .