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3 years ago

A set of four postcards costs $1.00. Single postcards cost $0.50. What is the least amount of money you

2 answers:

7 0

You would buy the set of four postcards 4 times. This would equal $4 for 12 postcards. Then you would have to buy 3 individual postcards. Those 3 postcards would be $1.50
$4+$1.50=$5.50

shepuryov [24]3 years ago

3 0

Answer: 3 Sets of the 4 post cards and 3 of the single post cart sets.

Step-by-step explanation:

You buy 3 of the 4 post card set becuase that is the most you can get of those to get under 15 post cards.

You would now have 12 postcards now you have to buy 3 of the single set to get to 15 cards

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Maurinko [17]

Answer:

(1,0.85)

Step-by-step explanation:

(x,y) = (t,d)

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The sum of three consecutive terms of an A.P is 18 and their product is 120. Find the terms?

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3 years ago

What is the area of the figure?

xenn [34]

Answer: 17.5 square yards

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Work Shown:

base = 5 yards

height = 7 yards

The base and height are always perpendicular to one another

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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.

Let $\bar X$ = sample mean heights

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)

= 0.99492 or 99.5%

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.

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.

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