What is the length of cd

A sweet factory produces 5 different chocolate bars. the different flavors are always produced in the same proportions, for ever

IRISSAK [1]

a. There are 933 chocolate bars

b. There are 1200 chocolate bars

c. There are 1580 chocolate bars

<h3>How to calculate the number of chocolate bars</h3>

Since there are 5 different flavors,

let

x = orange flavored bars,
y = coconut flavored bars,
z = coffee flavored bars,
a = strawberry flavored bars and
b = honeycomb flavored bars and
X = total number of bars

Since the different flavors are always produced in the same proportions, for every 4 coconut flavored bars 5 honeycomb 6 orange 1 coffee and 4 strawberry

So, the ratio of their proportions are x:y:z:a:b = 6:4:2:1:5

So, the total ratio is T = 6 + 4 + 1 + 4 + 5 = 20

<h3>a. What is the total number of chocolate bars if there are 280 orange flavored bars?</h3>

Since we have 6 orange flavored bars, the ratio of orange flavored bars to total is 6/20

So, the amount of orange flavored bars is x = 6/20 × X

Making X subject of the formula, we have

X = 20x/6

So, if there are 280 orange bars, there will be

X = 20x/6

X = 20 × 280/6

X = 5600/6

X = 933.33

X ≅ 933 chocolate bars

So, there are 933 chocolate bars

<h3>b. What is the total number of chocolate bars if there are 960 coconut flavored bars?</h3>

Since we have 4 coconut flavored bars, the ratio of coconut flavored bars to total is 4/20

So, the amount of coconut flavored bars is y = 4/20 × X

Making X subject of the formula, we have

X = 20y/4

So, if there are 960 coconut flavored bars, there will be

X = 20y/6

X = 20 × 960/4

X = 5 × 240

X = 1200 chocolate bars

So, there are 1200 chocolate bars

<h3>c. What is the total number of chocolate bars if there are 79 coffee flavored bars?</h3>

Since we have 1 coffee flavored bars, the ratio of coffee flavored bars to total is 1/20

So, the amount of coconut flavored bars is z = 1/20 × X

Making X subject of the formula, we have

X = 20z

So, if there are 79 coffee flavored bars, there will be

X = 20z

X = 20 × 79

X = 1580 chocolate bars

So, there are 1580 chocolate bars

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7 0

2 years ago

After once again losing a football game to the college'ss arch rival, the alumni association conducted a survey to see if alumni

antoniya [11.8K]

Answer:

a) The 90% confidence interval would be given (0.561;0.719).

b) $p_v =P(z>2.8)=1-P(z$

c) Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

Step-by-step explanation:

1) Data given and notation

n=100 represent the random sample taken

X=64 represent were in favor of firing the coach

$\hat p=\frac{64}{100}=0.64$ estimated proportion for were in favor of firing the coach

$p_o=0.5$ is the value that we want to test since the problem says majority

$\alpha$ represent the significance level (no given, but is assumed)

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

p= population proportion of Americans for were in favor of firing the coach

Part a

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 90% confidence interval the value of $\alpha=1-0.9=0.1$ and $\alpha/2=0.05$ , with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.64$

And replacing into the confidence interval formula we got:

$0.64 - 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.561$

$0.64 + 1.64 \sqrt{\frac{0.64(1-0.64)}{100}}=0.719$

And the 90% confidence interval would be given (0.561;0.719).

Part b

We need to conduct a hypothesis in order to test the claim that the proportion exceeds 50%(Majority). :

Null Hypothesis: $p \leq 0.5$

Alternative Hypothesis: $p >0.5$

We assume that the proportion follows a normal distribution.

This is a one tail upper test for the proportion of union membership.

The One-Sample Proportion Test is "used to assess whether a population proportion $\hat p$ is significantly (different,higher or less) from a hypothesized value $p_o$ ".

Check for the assumptions that he sample must satisfy in order to apply the test

a)The random sample needs to be representative: On this case the problem no mention about it but we can assume it.

b) The sample needs to be large enough

$np_o =100*0.64=64>10$

$n(1-p_o)=100*(1-0.64)=36>10$

Calculate the statistic

The statistic is calculated with the following formula:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o(1-p_o)}{n}}}$

On this case the value of $p_o=0.5$ is the value that we are testing and n = 100.

$z=\frac{0.64 -0.5}{\sqrt{\frac{0.5(1-0.5)}{100}}}=2.8$

The p value for the test would be:

$p_v =P(z>2.8)=1-P(z$

Part c

Using the significance level assumed $\alpha=0.01$ we see that $p_v$ so we have enough evidence at this significance level to reject the null hypothesis. And on this case makes sense the claim that the proportion is higher than 0.5 or 50%.

6 0

3 years ago

Is 3x^2-5x^4+2x-12 a polynomial

ivanzaharov [21]

Yes it is a polynomial .

The degree of polynomial is 4 and there is no power as a fraction or negative integer

6 0

3 years ago

Solve each problem and check your answer using estimates. a. $18.79 + $2.11 + ‐$1.92 + $17.28 b. $7.45 + ‐$24.45 + $74.17 + ‐$76

Aneli [31]

The solution to the linear expressions are:

a. $36.26
b. -$19.35
c. $70.38

<h3>Solving linear expressions:</h3>

The solution to linear expression is determined by taking into consideration the arithmetic operations used in each linear expression.

From the information given:

a. $18.79 + $2.11 + ‐$1.92 + $17.28

By rearrangement:

= $18.79 + $2.11 + $17.28 ‐$1.92

= $36.26

b. $7.45 + ‐$24.45 + $74.17 + ‐$76.52

By rearrangement:

= $7.45 + $74.17 ‐ $24.45 ‐ $76.52

= -$19.35

c. $98.45 − $10.63 + $2.82 − $20.26

By rearrangement:

= $98.45 + $2.82 − $10.63 − $20.26

= $70.38

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7 0

2 years ago

I don’t know when the two lines,rays, and segments are pls help

Katen [24]

Answer:

Lines- A, B

Segments- AC, FB

Rays- BD, DC, BC

5 0

3 years ago