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3 years ago

1. Clare has a recipe for yellow cake. She uses 9 1/3 cups of flour to make 4 cakes, Noah

Mathematics

1 answer:

Aleksandr [31]3 years ago

6 0

Answer: Relationship between c and f is 7c - 3f = 0

Since Clare has to use 9 cups each of them having 1/3 part filled with flour for making of 4 cakes.

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Speed compares two values in a ratio. calculate the average speed of karen who travels for 2 hours. karen traveled approximately

Mama L [17]

Speed=distance/time
distance=176km
time=2hr

speed=176km/2hr=88km/1hr

88kmph

5 0

3 years ago

At the city museum, child admission is $6.10 and adult admission is $9.90. On Friday, 142 tickets were sold for a total sales of

antoniya [11.8K]

Answer:

no of child tickets sold were 58

Step-by-step explanation:

Let the number of child ticket sold be x

cost of 1 child ticket = $6.10

Total cost of x child ticket = x*cost of 1 child ticket = = $6.10x

Let the number of adult ticket sold be y

cost of 1 adult ticket = $9.90

Total cost of y adult ticket = y*cost of 1 adult ticket = = $9.90y

Given total ticket sold = 142

Thus,

number of child ticket sold + number of adult ticket sold = 142

x + y = 142 - equation 1

x = 142 -y

Total sales of ticket = $1185.40

total sales of ticket = Total cost of x child ticket + Total cost of x a dult ticket

6.10x + 9.90y = 1185.40

in this equation substituting x = 142 - y from equation 1

6.10(142 - y ) + 9.90y = 1185.40

=>866.2 - 6.10y + 9.90y = 1185.40

=> 3.80y = 1185.40 - 866.2

=> 3.80y = 319.2

=> y = 319.2/3.80 = 84

Thus, y = 84

x = 142 - y = 142 - 84 = 58

Thus, no of child tickets sold were 58

6 0

3 years ago

Wal-Mart conducted a study to check the accuracy of checkout scanners at its stores. At each of the 60 randomly selected Wal-Mar

Mamont248 [21]

Answer:

The 95% confidence interval is

$0.7811 < p < 0.9529$

Generally the interval above can interpreted as

There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval

Generally 99% is outside the interval obtained in a above then the claim of Wal-mart is not believable

$n = 125 \ stores$

Step-by-step explanation:

From the question we are told that

The sample size is n = 60

The number of stores that had more than 2 items price incorrectly is k = 52

Generally the sample proportion is mathematically represented as

$\^ p = \frac{ k }{ n }$

=> $\^ p = \frac{ 52 }{ 60 }$

=> $\^ p = 0.867$

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{\frac{\^ p (1- \^ p)}{n} }$

=> $E = 1.96 * \sqrt{\frac{ 0.867 (1- 0.867)}{60} }$

=> $E = 0.0859$

Generally 95% confidence interval is mathematically represented as

$\^ p -E < p < \^ p +E$

=> $0.867 - 0.0859 < p < 0.867 + 0.0859$

=> $0.7811 < p < 0.9529$

Generally the interval above can interpreted as

There is 95% confidence that the true proportion of Wal-Mart stores that have more than 2 items priced inaccurately per 100 items scanned lie within the interval

Considering question b

Generally 99% is outside the interval obtained in a above then the claim of Wal-mart is not believable

Considering question c

From the question we are told that

The margin of error is E = 0.05

From the question we are told the confidence level is 95% , hence the level of significance is

$\alpha = (100 - 95 ) \%$

=> $\alpha = 0.05$

Generally from the normal distribution table the critical value of $\frac{\alpha }{2}$ is

$Z_{\frac{\alpha }{2} } = 1.645$