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OverLord2011 [107]
3 years ago
10

Solve. 4 x + 6 < − 6 Thankyou In advance :>

Mathematics
1 answer:
andrew-mc [135]3 years ago
6 0

Answer:

x < -3

Step-by-step explanation:

Step 1: Subtract 6 from both sides.

  • 4x + 6 - 6 < -6 - 6
  • 4x < -12

Step 2: Divide both sides by 4.

  • \frac{4x}{4} < \frac{-12}{4}
  • x < -3

Therefore, the answer is x < -3.

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A​ half-century ago, the mean height of women in a particular country in their 20s was 64.7 inches. Assume that the heights of​
Ainat [17]

Answer:

99.5% of all samples of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches.

Step-by-step explanation:

We are given that a half-century ago, the mean height of women in a particular country in their 20's was 64.7 inches. Assume that the heights of​ today's women in their 20's are approximately normally distributed with a standard deviation of 2.07 inches.

Also, a samples of 21 of​ today's women in their 20's have been taken.

<u><em /></u>

<u><em>Let </em></u>\bar X<u><em> = sample mean heights</em></u>

The z-score probability distribution for sample mean is given by;

                          Z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean height of women = 64.7 inches

            \sigma = standard deviation = 2.07 inches

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the sample of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches is given by = P(\bar X \geq 65.86 inches)

  P(\bar X \geq 65.86 inches) = P( \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } } \geq \frac{65.86-64.7}{\frac{2.07}{\sqrt{21} } } ) = P(Z \geq -2.57) = P(Z \leq 2.57)

                                                                        = <u>0.99492  or  99.5%</u>

<em>The above probability is calculated by looking at the value of x = 2.57 in the z table which has an area of 0.99492.</em>

<em />

Therefore, 99.5% of all samples of 21 of​ today's women in their 20's have mean heights of at least 65.86 ​inches.

3 0
3 years ago
Which line is parallel to the line that passes through the points (1, 7) and (-3, 4)? A. B. C. D.
Helga [31]

Answer:

hope this helps you

7 0
3 years ago
Are 7xy and 2yx like terms
Vanyuwa [196]

Answer: no, 7xy and 2xy are not like terms

3 0
3 years ago
Read 2 more answers
I’m trying to correct this problem but I don’t know how can someone help me
BartSMP [9]

Remark

It is not a straight line distance from the park to the mall. None of the answers give you that result. And if you know what displacement is, none of the answers are really displacement either. The distance is sort of a "as the crow flies." distance. There's a stop off in the middle of town.

Method

You need to use the Pythagorean Formula twice -- once from the park to the city Center and once from the city center to the mall.

Distance from the Park to the city center.

a = 3   [distance east]

b = 4  [distance south]

c = ??

c^2 = 3^2 + 4^2        Take the square root of both sides.

c = sqrt(3^2 + 4^2)

c = sqrt(9 + 16)          Add

c = sqrt(25)

c = 5

So the distance from the park to the city center is 5 miles

Distance from City center to the mall

a = 2 miles [distance east]

b = 2 miles [distance north]

c = ??

c^2 = a^2 + b^2   Substitute

c^2 = 2^2 + 2^2   Expand this.

c^2 = 4 + 4

c^2 = 8             Take the square root of both sides.

sqrt(c^2) = sqrt(8)

c = sqrt(8)          This is the result

c = 2.8

Answer

Total distance = 5 + 2.8 = 7.8

6 0
3 years ago
9/7-8/6= <br> Find the expresstions
Sergio039 [100]
Answer:
8/7 divided by 9/7=8/9
8 0
3 years ago
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