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2 years ago

A stalk of corn grew 42 inches in a week. Find the unit rate in inches per day

2 answers:

8 0

1 week = 7 days
42 inches per week

We want to find out how many inches grow per day.

42 / 7 = 6

So, everyday the corn grows 6 inches.

kozerog [31]2 years ago

7 0

To find the unit rate, we need to set the number of days to 1. Therefore, we will divide by 7. We also have to divide 42 by 7 to make sure the equation stays proportional. Using this knowledge, we can see that if the corn grew 42 inches in 7 days, it grows 6 inches per day.

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Students at a private liberal arts college are classified as being freshmen, sophomores, juniors or seniors, and also according

labwork [276]

The total number of possible classifications for the students of this college is found by multiplying 4 (which is the classification for the year level:freshman, sophomore, juniou, senior) and 2 (which is the number of sexes: female and male). So 4 x 2 = 8. There are eight possible classifications, which are:
(Male, Freshman)
(Male, Sophomore)
(Male, Junior)
(Male, Senior)
(Female, Freshman)
(Female, Sophomore)
(Female, Junior)
(Female,Senior)

5 0

2 years ago

Matthew ran 3/8 mile and then walked 7/10 mile. Which pair of fractions can he use to find how far he went in all? (A 15/40 and

Strike441 [17]

Before we could add these numbers, 3/8 and 7/10 need a common denominator. Both 8 and 10 go into 40.

8 goes into 40 five times
3/8= (3*5)/40 = 15/40

10 goes into 40 four times
7/10= (7*4)/40= 28/40

ANSWER: A) 15/40 and 28/40

Hope this helps! :)

7 0

2 years ago

Read 2 more answers

Joanna set a goal to drink more water daily. The number of ounces of water she drank each of the last seven days is shown below.

butalik [34]

(a) Data with the eight day's measurement.
Raw data: [60,58,64,64,68,50,57,82],
Sorted data: [50,57,58,60,64,64,68,82]
Sample size = 8 (even)
mean = 62.875
median = (60+64)/2 = 62
1st quartile = (57+58)/2 = 57.5
3rd quartile = (64+68)/2 = 66
IQR = 66 - 57.5 = 8.5

(b) Data without the eight day's measurement.
Raw data: [60,58,64,64,68,50,57]
Sorted data: [50,57,58,60,64,64,68]
Sample size = 7 (odd)
mean = 60.143
median = 60
1st quartile = 57
3rd quartile = 64
IQR = 64 -57 = 7

Answers:
1. The average is the same with or without the 8th day's data. FALSE
2. The median is the same with or without the 8th day's data. FALSE
3. The IQR decreases when the 8th day is included. FALSE
4. The IQR increases when the 8th day is included. TRUE
5. The median is higher when the 8th day is included. TRUE

8 0

2 years ago

Read 2 more answers

The grade point average (GPA) of the students at Lakeview High School is normally distributed with a mean of 3.1 and a standard

Morgarella [4.7K]

Approximately 1718 have a score within that range.

We calculate the z-score for each end of this spectrum:

z = (X-μ)/σ = (2.5-3.1)/0.3 = -0.6/0.3 = -2

Using a z-table (http://www.z-table.com) we see that the area to the left of, less than, this z-score is 0.0228.

For the upper end:
z = (3.7-3.1)/0.3 = 0.6/0.3 = 2

Using a z-table, we see that the area to the left of, less than, this z-score is 0.9772.

The probability between these is given by subtracting these:
0.9772 - 0.0228 = 0.9544.

This means the proportion of people that should fall between these is 0.9544:
0.9544*1800 = 1717.92 ≈ 1718

4 0

2 years ago

The average annual amount American households spend for daily transportation is $6312 (Money, August 2001). Assume that the amou

lions [1.4K]

Answer:

(a) The standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is 0.2283.

(c) The range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.

Step-by-step explanation:

We are given that the average annual amount American households spend on daily transportation is $6312 (Money, August 2001). Assume that the amount spent is normally distributed.

(a) It is stated that 5% of American households spend less than $1000 for daily transportation.

Let X = <u><em>the amount spent on daily transportation</em></u>

The z-score probability distribution for the normal distribution is given by;

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

where, $\mu$ = average annual amount American households spend on daily transportation = $6,312

$\sigma$ = standard deviation

Now, 5% of American households spend less than $1000 on daily transportation means that;

P(X < $1,000) = 0.05

P( $\frac{X-\mu}{\sigma}$ < $\frac{\$1000-\$6312}{\sigma}$ ) = 0.05

P(Z < $\frac{\$1000-\$6312}{\sigma}$ ) = 0.05

In the z-table, the critical value of z which represents the area of below 5% is given as -1.645, this means;

$\frac{\$1000-\$6312}{\sigma}=-1.645$

$\sigma=\frac{-\$5312}{-1.645}$ = 3229.18

So, the standard deviation of the amount spent is $3229.18.

(b) The probability that a household spends between $4000 and $6000 is given by = P($4000 < X < $6000)

P($4000 < X < $6000) = P(X < $6000) - P(X $\leq$ $4000)

P(X < $6000) = P( $\frac{X-\mu}{\sigma}$ < $\frac{\$6000-\$6312}{\$3229.18}$ ) = P(Z < -0.09) = 1 - P(Z $\leq$ 0.09)

= 1 - 0.5359 = 0.4641

P(X $\leq$ $4000) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{\$4000-\$6312}{\$3229.18}$ ) = P(Z $\leq$ -0.72) = 1 - P(Z < 0.72)

= 1 - 0.7642 = 0.2358

Therefore, P($4000 < X < $6000) = 0.4641 - 0.2358 = 0.2283.

(c) The range of spending for 3% of households with the highest daily transportation cost is given by;

P(X > x) = 0.03 {where x is the required range}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-\$6312}{3229.18}$ ) = 0.03

P(Z > $\frac{x-\$6312}{3229.18}$ ) = 0.03

In the z-table, the critical value of z which represents the area of top 3% is given as 1.88, this means;

$\frac{x-\$6312}{3229.18}=1.88$

${x-\$6312}=1.88\times 3229.18$

x = $6312 + 6070.86 = $12382.86

So, the range of spending for 3% of households with the highest daily transportation cost is $12382.86 or more.

8 0

2 years ago