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the perimeter of a rectangle is 55 inches. the ratio of the length to width is 7:4. use proportions to find the area of the rect

Leni [432]

Answer:

Step-by-step explanation:

let : x the length and y the width .... ( x > y and x ; y reals numbers )

the perimeter is : 2(x+y)

2(x+y) = 55...(1)

x/y = 7/4 ...(2)

by (2) : x = (7/4)y

subsct in (1) : 2( (7/4)y +y ) =55

(7/2)y +2y =55

multiply by 2

7y +4y = 110

11y = 110

y = 10

x = (7/4)(10)

x = 35/2

4 0

3 years ago

Rina is making a nut mixture to sell at the local farmers' market. She mixes 6 pounds of pecans with a nut mixture that is 70% p

Marat540 [252]

So the best way to do these is concentration1 (%) × volume1 = concentration2 × volume2
Or C1V1 + C2V2 = C3V3, where C1 = 100% (bc ALL pecans), V1 = 6 lbs, C2 = 70%, C3 = 82%:
100%×6 + 70%×v2 = 82%×(6+v2)
100%=1.00, 70%=.7, 82%=.82
note: if none is poured out then v3 = v1+v2
6 + .7v2 = .82 (6+v2)
6 + .7v2 = 4.92 + .82v2
6 + .7v2 -.7v2 = 4.92 + .82v2 -.7v2
6 = 4.92 + .12v2
6-4.92 = 4.92-4.92 + .12v2
1.08 = .12v2
.12v2/.12 = 1.08/.12
v2 = 9 lbs
that's only v2!!!

For the final poundage, we need v3:

v3 = 6 + v2 = 6 + 9 = 15 lbs

7 0

2 years ago

A manufacturer of tires wants to advertise a mileage interval that ex-cludes no more than 10% of the mileage on tires he sells.

Sergio039 [100]

Answer:

This mileage interval is from 30120 miles and higher.

Step-by-step explanation:

Problems of normally distributed samples can be 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:

All he knows is that, for a large number of tires tested, the mean mileage was 25,000 miles, and the standard deviation was 4000 miles. This means that $\mu = 25000, \sigma = 4000$ .

A manufacturer of tires wants to advertise a mileage interval that ex-cludes no more than 10% of the mileage on tires he sells. What interval wouldyou suggest?

The lower end of this interval is X when Z has a pvalue of 0.90. That is $Z = 1.28$ .