Which expression is equivalent to 4/5?

Mathematics

1 answer:

Sonja [21]2 years ago

7 0

Step-by-step explanation:

4/5=0.8

whatever expressions that equals 0.8

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a mint plant grows 7/8 in 1 week and 13/16 inch the next week. use a common denominator to write an equivalent fraction for each

MAVERICK [17]

Answer:

28/32 and 26/32

Step-by-step explanation:

A common denominator of 8 and 16 is 32.

7/8 = (7 * 4)/(8 * 4) = 28/32

13/16 = (13 * 2)/(16 * 2) = 26/32

Two equivalent fractions with a common denominator are 28/32 and 26/32.

4 0

3 years ago

Seperate the number 53 into 3 parts so that the second number is 3 less than the first number and the third number is twice the

irga5000 [103]

Hello,

Let a the first, b the second and c the third.
a+b+c=53
b=a/3
c=2a
==>a+a/3+2a=53
==>10a/3=53
==>a=15.9
c=2*a=2*15.9=31.8
b=a/3=15.9/3=5.3

Proof: a+b+c=15.9+5.3+31.8=53

The second number is 5.3

7 0

3 years ago

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)