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Zolol [24]
4 years ago
12

What does it Mean to have the same size shape and measure

Mathematics
1 answer:
kiruha [24]3 years ago
5 0

When something has the same size,shape and measure it is congruent

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A card is drawn at random out of a​ well-shuffled deck of 52 cards. Find the probability of drawing a six comma seven comma or e
Rina8888 [55]

Answer: \frac{3}{13}, or ~23.07%

Step-by-step explanation:

There are 52 cards in your average deck.

For this question, you need the numbers of the cards in the deck. There are four of each number from 2 through 10, and four aces, jacks, queens, and kings.

Considering there are four 6s, four 7s, and four 8s, there are 12 cards that fit the description.

Divide 12 by the total number of cards in the deck, 52, to get your answer.

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4 years ago
The graph shows how many cans of each type of cat food were sold one day.
hammer [34]
The answer would be C
4 0
3 years ago
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In an effort to estimate the mean of amount spent per customer for dinner at a major Lawrence restaurant, data were collected fo
larisa86 [58]

Answer:

The 99% confidence interval for the population mean is 22.96 to 26.64

Step-by-step explanation:

Consider the provided information,

A sample of 49 customers. Assume a population standard deviation of $5. If the sample mean is $24.80,

The confidence interval if 99%.

Thus, 1-α=0.99

α=0.01

Now we need to determine z_{\frac{\alpha}{2}}=z_{0.005}

Now by using z score table we find that  z_{\frac{\alpha}{2}}=2.58

The boundaries of the confidence interval are:

\mu-z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80-2.58\times \frac{5}{\sqrt{49}}=22.96\\\mu+z_{\frac{\alpha}{2}}\times \frac{\sigma}{\sqrt{n} }\\24.80+2.58\times \frac{5}{\sqrt{49}}=26.64

Hence, the 99% confidence interval for the population mean is 22.96 to 26.64

3 0
3 years ago
Which of the following defines liquid measure?
Aleks04 [339]
B units used to measure the volume of liquids such as water gas and milk because the measure of the liquid is defined by the unit used
6 0
3 years ago
If you can solve it.Thanks in advance
vivado [14]

Answer:

f(g(x))=\frac{1}{(x^{2}+1)^{2}} +\sqrt[3]{x^{2}+1}

Step-by-step explanation:

we have

f(x)=x^{2} +\frac{1}{\sqrt[3]{x}}

g(x)=\frac{1}{x^{2}+1}

we know that

In the function

f(g(x))

The variable of the function f is now the function g(x)

substitute

f(g(x))=(\frac{1}{x^{2}+1})^{2} +\frac{1}{\sqrt[3]{(\frac{1}{x^{2}+1})}}

f(g(x))=\frac{1}{(x^{2}+1)^{2}} +\sqrt[3]{x^{2}+1}

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