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MrRa [10]
3 years ago
14

Zach spent 2/3 hour reading on Friday and 1 1/3 hours reading on Saturday. How much more time did he read on Saturday than on Fr

iday?
Mathematics
2 answers:
hodyreva [135]3 years ago
7 0
Subtract.

1 1/3 - 2/3

Convert 1 1/3 into an improper fraction:

4/3 - 2/3

Subtract the numerators together:

2/3

So he read 2/3 more time on Saturday than on Friday.
11Alexandr11 [23.1K]3 years ago
3 0

Answer:  She read 2/3 hour more on Saturday than on Friday.

Step-by-step explanation:

Here, the number of hour she read on Friday =  \frac{2}{3}

While, the number of hour she read on Saturday =  1\frac{1}{3}=\frac{4}{3}

Hence, the difference between the number of hours she read on Saturday and Friday

=\frac{4}{3}-\frac{2}{3}

=\frac{4-2}{3}

=\frac{2}{3}

Hence, She read 2/3 hour more on Saturday than on Friday.

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Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters,
valentinak56 [21]

Answer:

The probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and appropriately huge random samples (n > 30) are selected from the population with replacement, then the distribution of the sample  means will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

\mu_{\bar x}=\mu

And the standard deviation of the distribution of sample mean  is given by,

\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}

The information provided is:

<em>μ</em> = 144 mm

<em>σ</em> = 7 mm

<em>n</em> = 50.

Since <em>n</em> = 50 > 30, the Central limit theorem can be applied to approximate the sampling distribution of sample mean.

\bar X\sim N(\mu_{\bar x}=144, \sigma_{\bar x}^{2}=0.98)

Compute the probability that the sample mean would differ from the population mean by more than 2.6 mm as follows:

P(\bar X-\mu_{\bar x}>2.6)=P(\frac{\bar X-\mu_{\bar x}}{\sigma_{\bar x}} >\frac{2.6}{\sqrt{0.98}})

                           =P(Z>2.63)\\=1-P(Z

*Use a <em>z</em>-table for the probability.

Thus, the probability that the sample mean would differ from the population mean by more than 2.6 mm is 0.0043.

8 0
3 years ago
Answer the screenshot, please.<br> (no links! PDFs are fine.)
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Answer:

Mode: 91

Median: 91

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2 years ago
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Arturiano [62]

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-7 and -5

Step-by-step explanation:

-7 and -5 added together are -12, and their product is 35.

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What is the answer to 7.26 x 106 ft​
Diano4ka-milaya [45]

Answer:

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Step-by-step explanation:

its like doing 726 times 106 then just move the decimal. and i looked it up

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Pre-image point N(6,-3) was dilated to point N'(2, -1). What was the scale factor used?
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Answer:

scale factor = \frac{1}{3}

Step-by-step explanation:

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