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3 years ago

The recipe for a Bach of brownies calls for 1/3 cup of oil. You only want to make 1/4 of a Bach . How much oil do you need

Mathematics

1 answer:

Zielflug [23.3K]3 years ago

5 0

Answer:

1 1/3

Step-by-step explanation:

1/3 divided by 1/4

keep change flip so 1/3 times 4/1 multiply straight across

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What is 3 2/5 x 2 1/3 ?

Hitman42 [59]

3 2/5 x 2 1/3 rewrite

12/5 x 7/3 convert into fractions instead of mixed fraction then multiply straight across

84/15 would be your final answer you might need to reduce it though :) hope i helped

5 0

3 years ago

Read 2 more answers

40% written as a decimal

Natasha2012 [34]

40%= 0.40 if you think of the decimal being at the end of the number (40.%) than just move the decimal twice makes it 0.40

8 0

3 years ago

Which point is a solution to the inequality

Nata [24]

Answer:

y < -x - 5

-4 < -(-3) - 5 = 3 - 5 = -2

So its point (-3, -4)

8 0

3 years ago

Researchers measured the data speeds for a particular smartphone carrier at 50 airports. The highest speed measured was 75.6 Mbp

olga55 [171]

Answer:

a) 59.98

b) 2.99

c) 2.99

d) Significantly High

Step-by-step explanation:

Part a)

Highest speed measured = x = 75.6 Mbps

Average/Mean speed = $\overline{x}$ = 15.62 Mbps

Standard Deviation = s = 20.03 Mbps

We need to find the difference between carrier's highest data speed and the mean of all 50 data speeds i.e. x - $\overline{x}$

x - $\overline{x}$ = 75.6 - 15.62 = 59.98 Mbps

Thus, the difference between carrier's highest data speed and the mean of all 50 data speeds is 59.98 Mbps

Part b)

In order to find how many standard deviations away is the difference found in previous part, we divide the difference by the value of standard deviation i.e.

$\frac{59.98}{20.03}=2.99$

This means, the difference is 2.99 standard deviations or in other words we can say, the Carrier's highest data speed is 2.99 standard deviations above the mean data speed.

Part c)

A z score tells us that how many standard deviations away is a value from the mean. We calculated the same in the previous part. Performing the same calculation in one step:

The formula for the z score is:

$z=\frac{x-\overline{x}}{s}$

Using the given values, we get:

$z=\frac{75.6-15.62}{20.03}=2.99$

Thus, the Carriers highest data is equivalent to a z score of 2.99

Part d)

The range of z scores which are neither significantly low nor significantly high is -2 to + 2. The z scores outside this range will be significant.

Since, the z score for carrier's highest data speed is 2.99 which is well outside the given range, i.e. greater than 2, we can conclude that the carrier's highest data speed is significantly higher.

3 0

3 years ago

The Information Technology Department at a large university wishes to estimate the proportion of students living in the dormitor

nadya68 [22]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16$

And rounded up we have that n=385

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by: