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3 years ago

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A box of Godiva chocolates contains 45 identically wrapped 8) chocolates. Ten are truffles, 21 have caramel fillings, the rest h

ave walnuts. What is the probability of a chocolate chosen at random of selecting a walnut?

2 answers:

zloy xaker [14]3 years ago

7 0

Answer:

14/45

Step-by-step explanation:

Probability is the likelihood or chance an event will occur.

Probability = Expected number of outcome/Total outcome

Total number of chocolate = Total outcome = 45

Before we can get the probability of a chocolate chosen at random of selecting a walnut, we need to know the number of chocolates that have walnuts.

Number of chocolates that have walnuts = 45-(10+21)

= 45-31

= 14

This means that our expected outcome is 14

Probability of a chocolate chosen at random of selecting a walnut = 14/45

I am Lyosha [343]3 years ago

6 0

Answer:

$\frac{7}{24}$

Step-by-step explanation:

A box of Godiva chocolates contains 45 identically wrapped chocolates. Ten are truffles, 21 have caramel fillings, the rest have walnuts.

Total number of chocolates = 45

total outcomes = 45

10 truffles, 21 caramel

number of walnuts = total chocolates - truffles - caramel

walnuts = 45 - 10 - 21= 14

Probability of selecting walnut = $\frac{number of walnut}{total chocolates}$

P(walnut) = $\frac{14}{45}$

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The diameters of bearings used in an aircraft landing gear assembly have a standard deviation of ???? = 0.0020 cm. A random samp

vovikov84 [41]

Answer:

(a) We conclude after testing that mean diameter is 8.2500 cm.

(b) P-value of test = 2 x 0.0005% = 1 x $10^{-5}$ .

(c) 95% confidence interval on the mean diameter = [8.2525 , 8.2545]

Step-by-step explanation:

We are given with the population standard deviation, $\sigma$ = 0.0020 cm

Sample Mean, Xbar = 8.2535 cm and Sample size, n = 15

(a) Let Null Hypothesis, $H_0$ : Mean Diameter, $\mu$ = 8.2500 cm

Alternate Hypothesis, $H_1$ : Mean Diameter, $\mu$ $\neq$ 8.2500 cm{Given two sided}

So, Test Statistics for testing this hypothesis is given by;

$\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ follows Standard Normal distribution

After putting each value, Test Statistics = $\frac{8.2535-8.2500}{\frac{0.0020}{\sqrt{15} } }$ = 6.778

Now we are given with the level of significance of 5% and at this level of significance our z score has a value of 1.96 as it is two sided alternative.

<em>Since our test statistics does not lie in the rejection region{value smaller than 1.96} as 6.778>1.96 so we have sufficient evidence to accept null hypothesis and conclude that the mean diameter is 8.2500 cm.</em>

(b) P-value is the exact % where test statistics lie.

For calculating P-value , our test statistics has a value of 6.778

So, P(Z > 6.778) = Since in the Z table the highest value for test statistics is given as 4.4172 and our test statistics has value higher than this so we conclude that P - value is smaller than 2 x 0.0005% { Here 2 is multiplied with the % value of 4.4172 because of two sided alternative hypothesis}

Hence P-value of test = 2 x 0.0005% = 1 x $10^{-5}$ .

(c) For constructing Two-sided confidence interval we know that:

Probability(-1.96 < N(0,1) < 1.96) = 0.95 { This indicates that at 5% level of significance our Z score will lie between area of -1.96 to 1.96.

P(-1.96 < $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96\frac{\sigma}{\sqrt{n} }$ < $Xbar - \mu$ < $1.96\frac{\sigma}{\sqrt{n} }$ ) = 0.95

P( $-Xbar-1.96\frac{\sigma}{\sqrt{n} }$ < $-\mu$ < $1.96\frac{\sigma}{\sqrt{n} }-Xbar$ ) = 0.95

P( $Xbar-1.96\frac{\sigma}{\sqrt{n} }$ < $\mu$ < $Xbar+1.96\frac{\sigma}{\sqrt{n} }$ ) = 0.95

So, 95% confidence interval for $\mu$ = [ $Xbar-1.96\frac{\sigma}{\sqrt{n} }$ , $Xbar+1.96\frac{\sigma}{\sqrt{n} }$ ]

= [ $8.2535-1.96\frac{0.0020}{\sqrt{15} }$ , $8.2535+1.96\frac{0.0020}{\sqrt{15} }$ ]

= [8.2525 , 8.2545]

Here $\mu$ = mean diameter.

Therefore, 95% two-sided confidence interval on the mean diameter

= [8.2525 , 8.2545] .

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fredd [130]

Answer:

a = 4,

b = 12

c = 10

d = 15

Step-by-step explanation:

Since the product of each column is equal, therefore,

b*5 = 60

b = 60 ÷ 5 = 12

c*6 = 60

c = 60 ÷ 6 = 10

Since the sum of each column are equal, therefore,

12 + 10 + a = 5 + 6 + d

22 + a = 11 + d

Think of a number you can add to 22, and another number you can add to 11, which will make both sides equal. Add both numbers, whenmultiplied together should give you 60.

Factors of 60 are:

(a, d)

(1, 60) => 22 + a = 11 + d => 22+1 = 11+60 (incorrect)

(2, 30) => 22 + a = 11 + d => 22+2 = 11+30 (incorrect)

(3, 20) => 22 + a = 11 + d => 22+3 = 21+20 (incorrect)

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(5, 12) => 22 + a = 11 + d => 22+5 = 11+12 (incorrect)

(6, 10) => 22 + a = 11 + d => 22+6 = 11+10 (incorrect)

Therefore,

a = 4,

d = 15

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MAXImum [283]

Answer:

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Step-by-step explanation:

The order of operations has to with a generally accepted order to follow when evaluating a mathematical expression.

PEDMAS or BODMAS is the order of operations to follow.

The two operations that are on the same level within the order of operations are addition and subtracttion

This means that, neither has a preference over the other.

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