All categories

4 years ago

Which expression represents the number of stamps Debbie has collected?

Mathematics

1 answer:

kvasek [131]4 years ago

3 0

Since she buys more, she adds them to the 53 she already has.
The expression is 53 + s

You might be interested in

Rectangular prism A measures 6 inches by 4 inches by 5 inches. Rectangular prism B's dimensions are twice those of prism A. Find

Rashid [163]

6×4×5 = 120 in^3; 12×8×10 = 960 in^3; Prism B's volume is 8 times greater than Prism A's volume

4 0

3 years ago

Read 2 more answers

What is the quoteint of 2/3 in 2/9

andre [41]

Answer: 3

Explanation:

2/3 divided by 2/9

2/3 times 9/2

= 3

6 0

4 years ago

Come quic 5 qts .. pics provided !! answer correctly plz!!

adelina 88 [10]

Answer:

First Picture: -2/7

Second Picture: 9/5

Third Picture: 5/3

Not sure of the others

Step-by-step explanation:

Pick two points on the line and determine their coordinates.

Determine the difference in y-coordinates of these two points (rise).

Determine the difference in x-coordinates for these two points (run).

Divide the difference in y-coordinates by the difference in x-coordinates (rise/run or slope).

7 0

3 years ago

Read 2 more answers

HELP ASAP PLEASE thanks youu

BlackZzzverrR [31]

Answer: Probably A

Step-by-step explanation: You need to use a ruler for this, because without one its impossible to get it exact

Using the ruler though, you'd measure each side of the wall and either use a ratio of 0.5in:4ft or multiply every 0.5in by 4ft.

6 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)

= 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.

3 0

3 years ago