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2 years ago

Melissa's chocolate bar is 77% cocoa. If the weight of the chocolate bar is 92 grams, how many grams of

Mathematics

2 answers:

Elena L [17]2 years ago

7 0

Answer:

70.8

Step-by-step explanation:

77% is the same as 0.77 so
92 * 0.77 = 77% of 92
92 * 0.77 = 70.84
rounded to the nearest tenth
70.8

mario62 [17]2 years ago

5 0

You can use the weight of the chocolate bar given and find its 77% to get the weight of cocoa used in it.

The weight of cocoa in the given chocolate bar is 70.80 grams.

<h3>How to find the percentage from the total value?</h3>

Suppose the value of which a thing is expressed in percentage is "a'

Suppose the percent that considered thing is of "a" is b%

Then since percent shows per 100 (since cent means 100), thus we will first divide the whole part in 100 parts and then we multiply it with b so that we collect b items per 100 items(that is exactly what b per cent means).

Thus, that thing in number is

$\frac{a}{100}$ × $b$

Since weight of cocoa = 77% percent of chocolate bar's weight which is 92 gram,

Thus,

$weight$ $of$ $cocoa=\frac{77}{100}$ × $92=68.08$ $grams$

Thus,

The weight of cocoa in the given chocolate bar is 70.80 grams.

Learn more about percentage here:

brainly.com/question/11549320

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valkas [14]

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$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

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We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

= 3 - 1 = 2

d.f.D = N - k

= 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

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$\overline X_1 =50.5$ $\overline X_2 =50.07143$ $\overline X_3 =55.71429$

$s_1^2=19.96154$ $s_2^2=14.68681$ $s_3^2=36.57143$

The grand mean or the overall mean is(GM) :

$\overline X_{GM}=\frac{\sum \overline X}{N}$

$=\frac{51+51+...+49+49}{35}$

= 52.1714

The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

$=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

$=\frac{127.1143}{2}$

= 63.55714

The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

$=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

$=\frac{669.8571}{32}$

= 20.93304

The F-test statistics value is :

$F=\frac{s_B^2}{s_W^2}$

$=\frac{63.55714}{20.93304}$

= 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source Sum of squares d.f Mean square F

Between 127.1143 2 63.55714 3.036212

Within 669.8571 32 20.93304

Total 796.9714 34

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2 years ago

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3 years ago

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