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3 years ago

PLEASE HELP FAST! WILL MARK AS BRAINLIEST IF GIVEN GOOD ANSWER!

Mathematics

1 answer:

soldier1979 [14.2K]3 years ago

5 0

Answer: If there is a lower average or score then the IQR's range will be lower. In this case, the lowest is around a 15 meaning that the IQR will move down, The 15 in this case is the outlier.

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The Spanish Club has 78 members, 42 girls and 36 boys. What is the ratio of

elena-14-01-66 [18.8K]

Answer:

C. 7:6

Step-by-step explanation:

Take your ratio 42 (G): 36 (B) and divide both numbers by six. This comes out to 7 (G): 6 (B) which is the simplest form of this ratio.

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2 years ago

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In ΔSTU, m∠S=(11x+3)∘,∠T=(2x+16)∘, and m∠U=(x+7)∘. Find .

frozen [14]

Answer:

x = 11

Step-by-step explanation:

In ΔSTU ,

m∠S = (11x+3)° ; m∠T = (2x+16)° ; m∠U = (x+7)°

According to angle sum property of a triangle , sum of all the interior angles of the triangle is 180°.

So,

$11x + 3 + 2x + 16 + x + 7 = 180$

$= > 14x + 26 = 180$

$= > 14x = 180 - 26 = 154$

$= > x = \frac{154}{14} = 11$

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3 years ago

Which expression is equivalent to 4^7 x 4^-5

melisa1 [442]

please mark this answer as brainlist

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3 years ago

Help!!!!!

photoshop1234 [79]

<span>c) 45 degrees west of south</span>

7 0

2 years ago

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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>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>

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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