38 ×29 use rounding to estimate each product

A candy store sells 8-pound bags of mixed hazelnuts and cashews. If c pounds of cashews are in a bag, the price p of the bag can

DochEvi [55]

A candy store sells 8 pound bags of mixed almonds and cashews. If c pounds of cashews are in a bag,...

c=1 was already given as described. The store sells pound bags.

The rest of the description and question seem not to match. Maybe each bag contains 8 pounds of the two nuts. Eight pounds and some of this is of cashews. Now, c is how many pounds of cashews in one bag.

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p=2.59%2Ac%2B1.72%288-c%29

You want to know the quantity of cashews if p=18.11 dollars per bag.

2.59c%2B1.72%2A8-1.72c=p

2.59c-1.72c=p-1.72%2A8

%282.59-1.72%29c=p-1.72%2A8

0.87c=p-13.76

7 0

2 years ago

Please help me understand, thank you.

monitta

Answer:

195.6 mph, 069 degrees

Step-by-step explanation:

two different ways, same result

7 0

2 years ago

If the measure of an angle is 13 degree, find measure 0f its supplement

amid [387]

Answer:

167 degrees

Step-by-step explanation:

supplementary angles equal to be 180 degrees, so subtract the given angle from 180 to find the missing angle.

180-13=167

5 0

3 years ago

Can you answer this? <br> factorise 6p + 24q

Talja [164]

Answer:

6(p+4q)

Step-by-step explanation:

6 0

3 years ago

The increasing annual cost (including tuition, room, board, books, and fees) to attend college has been widely discussed (Time).

NeX [460]

Answer:

(a) PRIVATE COLLEGES

Sample mean is $42.5 thousand

Sample standard deviation is $6.65 thousand

PUBLIC COLLEGES

Sample mean is $22.3 thousand

Sample standard deviation is $4.34 thousand

(b) Point estimate is $20.2 thousand. The mean annual cost to attend private colleges ($42.5 thousand) is more than the mean annual cost to attend public colleges ($22.3 thousand)

(c) 95% confidence interval of the difference between the mean annual cost of attending private and public colleges is $19.2 thousand to $21.2 thousand

Step-by-step explanation:

(a) PRIVATE COLLEGES

Sample mean = Total cost ÷ number of colleges = (51.8+42.2+45+34.3+44+29.6+46.8+36.8+51.5+43) ÷ 10 = 425 ÷ 10 = $42.5 thousand

Sample standard deviation = sqrt[summation (cost - sample mean)^2 ÷ number of colleges] = sqrt([(51.8-42.5)^2 + (42.2-42.5)^2 + (45-42.5)^2 + (34.3-42.5)^2 + (44-42.5)^2 + (29.6-42.5)^2 + (36.8-42.5)^2 + (51.5-42.5)^2 + (43-42.5)^2] ÷ 10) = sqrt (44.24) = $6.65 thousand

PUBLIC COLLEGES

Sample mean = (20.3+22+28.2+15.6+24.1+28.5+22.8+25.8+18.5+25.6+14.4+21.8) ÷ 12 = 267.6 ÷ 12 = $22.3 thousand

Sample standard deviation = sqrt([(20.3-22.3)^2 + (22-22.3)^2 + (28.2-22.3)^2 + (15.6-22.3)^2 + (24.1-22.3)^2 + (28.5-22.3)^2 + (22.8-22.3)^2 + (25.8-22.3)^2 + (18.5-22.3)^2 + (25.6-22.3)^2 + (14.4-22.3)^2 + (21.8-22.3)^2] ÷ 12) = sqrt (18.83) = $4.34 thousand

(b) Point estimate = mean annual cost of attending private colleges - mean annual cost of attending public colleges = $42.5 thousand - $22.3 thousand = $20.2 thousand.

This implies the the mean annual cost of attending private colleges is greater than the mean annual cost of attending public colleges

(c) Confidence Interval = Mean + or - Margin of error (E)

E = t×sd/√n

Mean = $42.5 - $22.3 = $20.2 thousand

sd = $6.65 - $4.34 = $2.31 thousand

n = 10+12 = 22

degree of freedom = 22-2 = 20

t-value corresponding to 20 degrees of freedom and 95% confidence level is 2.086

E = 2.086×$2.31/√22 = $1.0 thousand

Lower bound = Mean - E = $20.2 thousand - $1.0 thousand = $19.2 thousand

Upper bound = Mean + E = $20.2 thousand + $1.0 thousand = $21.2 thousand

95% confidence interval is $19.2 thousand to $21.2 thousand

6 0

2 years ago