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3 years ago

5

a brownie recipe requires two and two thirds cups of sugar to one cup of chocolate chips. If one and one third cups of sugar is

used, what quantity of chocolate chips will be needed, according to the recipe

1 answer:

Taya2010 [7]3 years ago

4 0

Half of what the recipe required because your using half the required amount of sugar

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Sam needs 7 1/2 cups of orange juice to make punch for a group of her friends. She only has 5 1/3 cups. Write and solve an equat

Solnce55 [7]

Answer:

Equation is $x+\frac{16}{3}=\frac{15}{2}$ where $x$ denotes number of cups of orange juice needed by Sam.

$x=\frac{13}{6}$

Step-by-step explanation:

Let $x$ denotes number of cups of orange juice needed by Sam.

Number of cups needed to make punch $=7\frac{1}{2} =\frac{15}{2}$

Number of cups with Sam $=5\frac{1}{3}=\frac{16}{3}$

Therefore,

$x+\frac{16}{3}=\frac{15}{2}\\\\x= \frac{15}{2}-\frac{16}{3}\\\\x=\frac{45-32}{6}\\\\x=\frac{13}{6}$

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2 years ago

Hat is the value of x in the equation −x = 3 − 4x + 15? −18 −6 6 18

nadya68 [22]

I believe that x is equal to 6. x=6

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1 year ago

In the chart of accounts, each account number has two digits. The first digit indicates the major account group to which the acc

Eduardwww [97]

Answer:

The correct option is (b)

Step-by-step explanation:

Chart of accounts refers to listing or arranging various accounts for the ease of locating them. Listing is done based on the order of appearance beginning with balance sheet and then income statement.

The order starts with assets, followed by liabilities and stockholders' equity from the balance sheet and revenue and expenses from income statement.

So, the correct order is stated in option (b).

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3 years ago

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A sample of size 6 will be drawn from a normal population with mean 61 and standard deviation 14. Use the TI-84 Plus calculator.

AVprozaik [17]

Answer:

$X \sim N(61,14)$

Where $\mu=61$ and $\sigma=14$

Since the distribution for X is normal then the distribution for the sample mean is also normal and given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\mu_{\bar X} = 61$

$\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715$

So then is appropiate use the normal distribution to find the probabilities for $\bar X$

Step-by-step explanation:

Previous concepts

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".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean". The letter $\phi(b)$ is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: $\phi(b)=P(z$

Solution to the problem

Let X the random variable that represent the variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(61,14)$

Where $\mu=61$ and $\sigma=14$

Since the distribution for X is normal then the distribution for the sample mean $\bar X$ is also normal and given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$\mu_{\bar X} = 61$

$\sigma_{\bar X}= \frac{14}{\sqrt{6}}= 5.715$

So then is appropiate use the normal distribution to find the probabilities for $\bar X$

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3 years ago

A class has 14 girls and 12 boys. What is the probability that a girl's name is drawn at random?

nirvana33 [79]

The answer will be B

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3 years ago

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