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katen-ka-za [31]
3 years ago
7

PLEASE HELP!!!!!! Find the image of (1,2) after a reflection about x=6 followed by a reflection about x=4

Mathematics
1 answer:
Bond [772]3 years ago
5 0
I would help you but it me hard for me too
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Which is greater, - 5¼ or -5⅓
lana [24]

Answer:

-51/4 is greater

Step-by-step explanation:

7 0
3 years ago
Suppose x represents an objects weight on Earth. Write two expressions: one that you can use to find the objects weight on Mars
tatyana61 [14]
The weight of an object is the product of its mass and the acceleration of gravity.
If g[e] is the acceleration of gravity on earth, and g[M] the same for Mars and g[m] the same for the moon,
then m[M]=m[e]g[M]/g[e] and m[m]=m[e]g[m]/g[e] where m[ ] denotes mass. Note that weight=mg (measured in newtons) while mass is in kilograms.
If g[M]=g[e]/3 and g[m]=g[e]/6 approximately. Then the weight of an object on Mars will be about a third of what it is on earth, while on the moon it would be about a sixth of what it is on earth.
4 0
3 years ago
Please help me and we should use laws of exponents​
elena55 [62]

Answer:

3.b

(\frac{8}{3})^{2x+1} *(\frac{8}{3})^{5} =(\frac{8}{3}) ^{x+2}\\\\(\frac{8}{3})^{2x+1 +5} =(\frac{8}{3}) ^{x+2}\\\\\\2x +6 = x+2\\x = -4

i hope question 3.a. is 5^x ÷ 5^-3 = 5^5

3.a.

\frac{5^x}{5^{-3}} = 5^5\\\\5^ {x+3} = 5^5\\\\x+3 = 5\\x =2

6 0
3 years ago
PLZ ANSWER ASAP (EASY)
Anuta_ua [19.1K]

Answer:

-4/8

Step-by-step explanation:

that's what my math says

4 0
3 years ago
The lifespan (in days) of the common housefly is best modeled using a normal curve having mean 22 days and standard deviation 5.
Natasha_Volkova [10]

Answer:

Yes, it would be unusual.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are 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.

If Z \leq -2 or Z \geq 2, the outcome X is considered unusual.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 22, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1

Would it be unusual for this sample mean to be less than 19 days?

We have to find Z when X = 19. So

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

By the Central Limit Theorem

Z = \frac{19 - 22}{1}

Z = -3

Z = -3 \leq -2, so yes, the sample mean being less than 19 days would be considered an unusual outcome.

7 0
3 years ago
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