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shtirl [24]
3 years ago
14

BRAINLISEST TO FIRST CORRECT ANSWER

Mathematics
1 answer:
Korolek [52]3 years ago
7 0

Answer:

24 rolls

Step-by-step explanation:

1\frac{1}{2} =\frac{3}{2}

\frac{3}{2} ×\frac{4}{1} = \frac{12}{2}

which is 6 cups of flour

rolls can be made with \frac{1}{4} a cup of flour

6 × 4 =24

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andre [41]

Answer:

x ≥ 1  (how to graph is listed below)

Step-by-step explanation:

To find where we need to plot the line, we first need to solve the inequality for x:

-2x - 3 ≤ -5

(Add three to both sides)

-2x ≤ -2

(Divide both sides by -2, but we can't forget that whenever we multiple or divide by a negative number, the sign flips!)

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To graph this on the number line, you would put a dot on the 1 and fill it in completely (you fill in the dot for a "___ and equal to" sign.     ex. ≥, ≤)

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8 0
3 years ago
A population of values has a normal distribution with μ=204.9μ=204.9 and σ=81.9σ=81.9. You intend to draw a random sample of siz
marin [14]

Answer:

7. μ=204.9 and σ=5.4968

8. μ=75.9 and σ=0.7136

9. p=0.9452

Step-by-step explanation:

7. - Given that the population mean =204.9 and the standard deviation is 81.90 and the sample size n=222.

-The sample mean,\mu_xis calculated as:

\mu_x=\mu=204.9, \mu_x=sample \ mean

-The standard deviation,\sigma_x is calculated as:

\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{81.9}{\sqrt{222}}\\\\=5.4968

8. For a random variable X.

-Given a X's population mean is 75.9, standard deviation is 9.6 and a sample size of 181

-#The sample mean,\mu_x is calculated as:

\mu_x=\mu\\\\=75.9

#The sample standard deviation is calculated as follows:

\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{9.6}{\sqrt{181}}\\\\=0.7136

9. Given the population mean, μ=135.7 and σ=88 and n=59

#We calculate the sample mean;

\mu_x=\mu=135.7

#Sample standard deviation:

\sigma_x=\frac{\sigma}{\sqrt{n}}\\\\=\frac{88}{\sqrt{59}}\\\\=11.4566

#The sample size, n=59 is at least 30, so we apply Central Limit Theorem:

P(\bar X>117.4)=P(Z>\frac{117.4-\mu_{\bar x}}{\sigma_x})\\\\=P(Z>\frac{117.4-135.7}{11.4566})\\\\=P(Z>-1.5973)\\\\=1-0.05480 \\\\=0.9452

Hence, the probability of a random sample's mean being greater than 117.4 is 0.9452

7 0
3 years ago
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