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Fantom [35]
3 years ago
13

after adjusting the number of servings in a recipe Charlotte determine that she will need 53 tbsp of flour one cup is equivalent

to 16T . how much cups of flour will be required​
Mathematics
1 answer:
MrRissso [65]3 years ago
3 0

Answer:

3.31 cups

Step-by-step explanation:

1/× =16/53

16x = 53

x = 53/16

x = 3.3125

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8 0
3 years ago
An accounting firm is planning for the next tax preparation season. From last years returns, the firm collects a systematic rand
Elena L [17]

Answer:

a)From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

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

And the standard error for the mean would be:

\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14

b) We want this probability:

P(\bar X >120)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

P( z>2.143) = 1-P(Z

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 central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Part a

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

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

And the standard error for the mean would be:

\sigma_{\bar X}= \frac{140}{\sqrt{100}} =14

Part b

We want this probability:

P(\bar X >120)

And we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And replacing we got:

z = \frac{120-90}{\frac{140}{\sqrt{100}}}= 2.143

And we can find this probability with the complement rule and the normal standard deviation or excel and we got:

P( z>2.143) = 1-P(Z

4 0
3 years ago
This is for the person thats been helping me thanks (:
sergeinik [125]

------- cant see image

7 0
3 years ago
Read 2 more answers
What's the measure of m∠JKL?
AleksandrR [38]

Answer:

sorry, I think yr question is incomplete.

3 0
3 years ago
together,sarah's two cats have the same mass as her dog,11 kg.the mass of one cat is 1500 g greater than the mass of the other.
Alika [10]
Let (a) be the first cat
let (b) be the second cat
a +b=11kg-----(1)
a=b+1500g⇒(1.5 kg)-------(2)
so now u have a system of two unknowns
a=11-b
substitute a in (2)
11-b=b+1.5
11-1.5=2b
2b=9.5
b=9.5/2=4.75
subs. b in (1)
a+b=11
a+4.75=11
a=11-4.75=6.25
so cat (a) is 6.25 kg
and (b) is 4.75 kg
hope I helped



6 0
3 years ago
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