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Umnica [9.8K]
3 years ago
5

What is 54.873015873 rounded to the nearest hundereth please help

Mathematics
2 answers:
SpyIntel [72]3 years ago
7 0

Answer:

54.87

Step-by-step explanation:

Ulleksa [173]3 years ago
3 0
The answer is 54.87 as it is the hundreth
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Would appreciate some help thanks
defon

Answer:

f(8) = -9

Step-by-step explanation:

Substitute 8 in for x.

f(8) = -2(8) + 7

f(8) = -16 + 7

f(8) = -9

6 0
3 years ago
Jenny has some flowers in a bucket. She picks a flower at
saul85 [17]

Answer:

Each flower is either red or yellow. Each flower is either a tulip or a rose. For these flowers number of tulips:number of roses 6 : 5.

Step-by-step explanation:

6 0
3 years ago
Please help!!! Thank you!!!
Reptile [31]
To find the total area of this figure, it would be easiest to find the area of the left part (rectangle) and then find the area of the right part (triangle), and then add the two area values together.

First, we will find the area of the rectangle, using the formula A = lw, where l is the length of the rectangle and w is the width of the rectangle.
The length of the rectangle is 13 cm and the width is 9 cm. If we substitute in these values into our equation, we get:
A = (13cm)(9cm)
A= 117 cm^2

Next, let’s find the area of the triangle, using the formula A=(1/2)bh, where b is the base of the triangle and h is the height.
The base of the triangle is 11 cm and the height of the triangle is 5 cm (found by subtracting 13-8 as seen in the figure). If we substitute in these values and simplify, we get:
A=1/2(11cm)(5cm)
A=1/2(55cm^2)
A=27.5 cm^2.

When we add together the area of the rectangle with the area of the triangle, we will get the total area of the figure.

117 cm^2 + 27.5 cm^2 = 144.5 cm^2
Your answer is 144.5 cm^2 or the first option.

Hope this helps!
8 0
3 years ago
The total amount of fiber (in grams) in a package containing x apples and y oranges is given by the equation 5x + 10y = 110. Is
Flauer [41]
The answer is NO
Because if x=5 the equation is 125+10y=110
In this case y will be a negative
You can’t have a negative oranges
So the answer is NO
3 0
3 years ago
Read 2 more answers
A random sample of n = 64 observations is drawn from a population with a mean equal to 20 and standard deviation equal to 16. (G
dezoksy [38]

Answer:

a) The mean of a sampling distribution of \\ \overline{x} is \\ \mu_{\overline{x}} = \mu = 20. The standard deviation is \\ \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{64}}=2.

b) The standard normal z-score corresponding to a value of \\ \overline{x} = 16 is \\ Z = -2.

c) The standard normal z-score corresponding to a value of \\ \overline{x} = 23 is \\ Z = 1.5.

d) The probability \\ P(\overline{x}.

e) The probability \\ P(\overline{x}>23) = 1 - P(Z.

f)  \\ P(16 < \overline{x} < 23) = P(-2 < Z < 1.5) = P(Z.

Step-by-step explanation:

We are dealing here with the concept of <em>a sampling distribution</em>, that is, the distribution of the sample means \\ \overline{x}.

We know that for this kind of distribution we need, at least, that the sample size must be \\ n \geq 30 observations, to establish that:

\\ \overline{x} \sim N(\mu, \frac{\sigma}{\sqrt{n}})

In words, the distribution of the sample means follows, approximately, a <em>normal distribution</em> with mean, \mu, and standard deviation (called <em>standard error</em>), \\ \frac{\sigma}{\sqrt{n}}.

The number of observations is n = 64.

We need also to remember that the random variable Z follows a <em>standard normal distribution</em> with \\ \mu = 0 and \\ \sigma = 1.

\\ Z \sim N(0, 1)

The variable Z is

\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}} [1]

With all this information, we can solve the questions.

Part a

The mean of a sampling distribution of \\ \overline{x} is the population mean \\ \mu = 20 or \\ \mu_{\overline{x}} = \mu = 20.

The standard deviation is the population standard deviation \\ \sigma = 16 divided by the root square of n, that is, the number of observations of the sample. Thus, \\ \frac{\sigma}{\sqrt{n}} = \frac{16}{\sqrt{64}}=2.

Part b

We are dealing here with a <em>random sample</em>. The z-score for the sampling distribution of \\ \overline{x} is given by [1]. Then

\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ Z = \frac{16 - 20}{\frac{16}{\sqrt{64}}}

\\ Z = \frac{-4}{\frac{16}{8}}

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

\\ Z = -2

Then, the <em>standard normal z-score</em> corresponding to a value of \\ \overline{x} = 16 is \\ Z = -2.

Part c

We can follow the same procedure as before. Then

\\ Z = \frac{\overline{x} - \mu}{\frac{\sigma}{\sqrt{n}}}

\\ Z = \frac{23 - 20}{\frac{16}{\sqrt{64}}}

\\ Z = \frac{3}{\frac{16}{8}}

\\ Z = \frac{3}{2}

\\ Z = 1.5

As a result, the <em>standard normal z-score</em> corresponding to a value of \\ \overline{x} = 23 is \\ Z = 1.5.

Part d

Since we know from [1] that the random variable follows a <em>standard normal distribution</em>, we can consult the <em>cumulative standard normal table</em> for the corresponding \\ \overline{x} already calculated. This table is available in Statistics textbooks and on the Internet. We can also use statistical packages and even spreadsheets or calculators to find this probability.

The corresponding value is Z = -2, that is, it is <em>two standard units</em> <em>below</em> the mean (because of the <em>negative</em> value). Then, consulting the mentioned table, the corresponding cumulative probability for Z = -2 is \\ P(Z.

Therefore, the probability \\ P(\overline{x}.

Part e

We can follow a similar way than the previous step.

\\ P(\overline{x} > 23) = P(Z > 1.5)

For \\ P(Z > 1.5) using the <em>cumulative standard normal table</em>, we can find this probability knowing that

\\ P(Z1.5) = 1

\\ P(Z>1.5) = 1 - P(Z

Thus

\\ P(Z>1.5) = 1 - 0.9332

\\ P(Z>1.5) = 0.0668

Therefore, the probability \\ P(\overline{x}>23) = 1 - P(Z.

Part f

This probability is \\ P(\overline{x} > 16) and \\ P(\overline{x} < 23).

For finding this, we need to subtract the cumulative probabilities for \\ P(\overline{x} < 16) and \\ P(\overline{x} < 23)

Using the previous <em>standardized values</em> for them, we have from <em>Part d</em>:

\\ P(\overline{x}

We know from <em>Part e</em> that

\\ P(\overline{x} > 23) = P(Z>1.5) = 1 - P(Z

\\ P(\overline{x} < 23) = P(Z1.5)

\\ P(\overline{x} < 23) = P(Z

\\ P(\overline{x} < 23) = P(Z

Therefore, \\ P(16 < \overline{x} < 23) = P(-2 < Z < 1.5) = P(Z.

5 0
3 years ago
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