Mathematics, 16.07.2019 22:00 Naikayah

Pennies made before 1983 are 97% copper and 3% zinc, whereas pennies made after 1983 are 3% copper and 97% zinc. listed below are the weights (in grams) of pennies from each of the two time periods. find the mean and median for each of the two samples, then compare the two sets of results.

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The table in the attached figure

Pennies made before 1983
Part a) find the mean
we have
[3.1047,3.1082, 3.1582, 3.0776, 3.1276, 3.1035]
Mean 1=[3.1047+3.1082 + 3.1582 + 3.0776 + 3.1276 + 3.1035]/6
Mean 1=3.1133 grams

the answer part a) is
The mean weight of the pennies made before 1983 is 3.1133 grams.

Part b)
to find the Median, place the numbers in value order and find the middle.
[3.1047,3.1082, 3.1582, 3.0776, 3.1276, 3.1035]
[3.0776, 3.1035, 3.1047,3.1082, 3.1276, 3.1582]
(3.1047+3.1082)/2=3.10645 grams

the answer part b) is
The median weight of the pennies made before 1983 is 3.10645 grams.

Pennies made after 1983
Part c) find the mean
we have
[2.4842, 2.4828, 2.5024, 2.4997, 2.4907, 2.4956]
Mean 2=[2.4842 + 2.4828 + 2.5024 + 2.4997 + 2.4907 + 2.4956]/6
Mean 2=2.4926 grams

the answer part c) is
The mean weight of the pennies made after 1983 is 2.4926 grams.

Part d)
to find the Median, place the numbers in value order and find the middle
[2.4842, 2.4828, 2.5024, 2.4997, 2.4907, 2.4956]
[2.4828, 2.4842, 2.4907, 2.4956, 2.4997, 2.5024]
[2.4907 +2.4956]/2=2.49315

the answer part d) is
The median weight of the pennies made after 1983 is 2.49315 grams.

Part e) compare the means
Find the percentage difference
the percentage difference is the difference between the new value and the old value, divided by the old value
so
new value=2.4926 grams
old value=3.1133 grams
[3.1133-2.4926]/3.1133------> 0.20-----> 20%
that means
the difference in the means is considerable to be greater than 5%

Pennies made before 1983 are 97% copper and 3% zinc, whereas pennies made after 1983 are 3% copper a

Pennies made before 1983 are 97% copper and 3% zinc, whereas pennies made after 1983 are 3% copper a

Answer from: Quest

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.

Point z is equidistant from the sides of δrst. which must be true? ctzaszaszzsb

Answer from: Quest

so, converting firstly the mixed fractions to improper fractions and then "adding".

$\bf \stackrel{mixed}{2\frac{9}{10}}\implies \cfrac{2\cdot 10+9}{10}\implies \stackrel{improper}{\cfrac{29}{10}}~\hfill \stackrel{mixed}{5\frac{4}{15}}\implies \cfrac{5\cdot 15+4}{15}\implies \stackrel{improper}{\cfrac{79}{15}} [-0.35em] ~\dotfill$

$\bf -\cfrac{29}{10}+\left( -\cfrac{79}{15} \right)\implies -\cfrac{29}{10}-\cfrac{79}{15}\implies \stackrel{\textit{using the lcd of 30}}{\cfrac{-(3)29-(2)79}{30}}\implies \cfrac{-87-158}{30} \cfrac{-245}{30}\implies -\cfrac{49}{6}\implies -8\frac{1}{6}$

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