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Mariulka [41]
2 years ago
12

Fill in the missing numbers in these equation (-2) . (-4.5)= ? ?=

Mathematics
1 answer:
Nina [5.8K]2 years ago
5 0
9
Multiplying two negatives makes it positive.
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I struggle a lot with pre-algebra pls help
ch4aika [34]

Answer:

Polygon B is similar to Polygon A.

Step-by-step explanation:

Polygon B has a <em>scale factor</em> of 2, because Polygon B looks identical to A, except the measurements are twice as large.

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2 years ago
2. Multiple Choice. Gary says, “I chose a number. I multiplied it by 5 and then
kkurt [141]

Answer:

The answer would be B

4 0
3 years ago
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What is three different ways to write 5^11 power as the product of two different powers please help.
Veronika [31]

Answer:

Three ways to write 5^11

In product:48828125

In powers:

5^{11}

5 to the power of 11

Step-by-step explanation:

I learned this before in class like few years ago

4 0
2 years ago
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1/5x-2/3y=30 when y=15
dolphi86 [110]

Answer:

Step-by-step explanation:

<u>Step 1:  Set y to 15 and solve</u>

<u />\frac{1}{5}x - \frac{2}{3}*15=30

\frac{1}{5}x-10=30

<em>Add 10 to both sides</em>

<em />\frac{1}{5}x-10+10=30+10

\frac{1}{5}x=40

<em>Multiply both sides by 5</em>

<em />\frac{1}{5}x*\frac{5}{1}=40*\frac{5}{1}

x=200

Answer:  x=200

5 0
3 years ago
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
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