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azamat
3 years ago
6

a farmer grows only corn and lettuce. the farmer plans to plant 455 rows this year. the number of rows of corn will be 2.5 times

the number of rows of lettuce. how many rows of each vegetable does the farmer plan to plant?
Mathematics
1 answer:
zepelin [54]3 years ago
7 0
2.5x + x = 455
3.5x = 455
455/3.5 = x
x = 130

corn : 325
lettuce: 130
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You bought a can of black beans last week that cost $1.34. This week, the store advertised a sale where 5 cans of beans cost $6.
Sonja [21]

Answer:

1. $1.2

2.  $0.14

Step-by-step explanation:

Step one:

given data

Black beans last week that cost $1.34

Step two:

This week's price

5 cans of beans cost $6.00

<u>The cost per can is</u>

= 6/5

=$1.2

<u>The difference in price</u>

=1.34-1.2

= $0.14

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3 years ago
Sam brought brushes for $8,a palette for $5,and oil paints for $15.He paid 29.82 in all.What sale-tax rate did Sam pay?
sertanlavr [38]
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5 0
3 years ago
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kramer

Answer:

V=144\ cm^3

Step-by-step explanation:

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h=\sqrt{13^2-(4^2+3^2)}\\\\h=12\ cm

#The volume of the beam can then be calculated as:

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5 0
3 years ago
I. In a shipment of 300 connecting rods, the mean tensile strength is found to be 45 kpsi and has a standard deviation of 5 kpsi
CaHeK987 [17]

Answer:

a) Between 39 and 40 rods can be expected to have a strength less than 39.4 kpsi.

b) 260 rods are expected to have a strength between 39.4 and 60 kpsi

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value 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. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 45, \sigma = 5

(a) Assuming a normal distribution, how many rods can be expected to have a strength less than 39.4 kpsi?

The percentage of rods with a stength less than 39.4 is the pvalue of Z when X = 39.4. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{39.4 - 45}{5}

Z = -1.12

Z = -1.12 has a pvalue of 0.1314

13.14% of rods have a strength less than 39.4 kpsi.

Out of 300

0.1314*300 = 39.42

Between 39 and 40 rods can be expected to have a strength less than 39.4 kpsi.

(b) How many are expected to have a strength between 39.4 and 60 kpsi?

The percentage of rods with a stength in this interval is the pvalue of Z when X = 60 subtracted by the pvalue of Z when X = 39.4. So

X = 60

Z = \frac{X - \mu}{\sigma}

Z = \frac{60 - 45}{5}

Z = 3

Z = 3 has a pvalue of 0.9987

X = 39.4

Z = \frac{X - \mu}{\sigma}

Z = \frac{39.4 - 45}{5}

Z = -1.12

Z = -1.12 has a pvalue of 0.1314

0.9987 - 0.1314 = 0.8673

86.73% of the rods are expected to have a strength between 39.4 and 60 kpsi

Out of 300

0.8673*300 = 260

260 rods are expected to have a strength between 39.4 and 60 kpsi

4 0
4 years ago
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nevsk [136]
So i believe the answer is d) 1/9
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