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Mama L [17]
3 years ago
5

Estimate 19.41 - 6.254 by first rounding each number to the nearest tenth.

Mathematics
1 answer:
nordsb [41]3 years ago
4 0

Answer:

13.1

Step-by-step explanation:

19.41 => 19.4

6.254 => 6.3

so, 19.4 - 6.3 = 13.1

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What is the value of 1/3x (2) + 2 when x=3
Svetlanka [38]
Final answer: 5
Step by step process:
You can make the equation a little simpler by putting in the 3 for x
btw I show exponents with this ^
(3)^2/3 + 2

3^2=3*3=9

Now you have 9/3 + 2 and you keep simplifying

9/3=3
3+2=5
Final answer: 5
8 0
3 years ago
4.
Sholpan [36]

Answer:

A. {(5, –4), (–4, 5), (–5, 4), (4, –5)}

Step-by-step explanation:

Since there is one value of  

y

for every value of  

x

in  

(

5

,

−

4

)

,

(

−

4

,

5

)

,

(

−

5

,

4

)

,

(

4

,

−

5

)

, this relation is a function.

The relation is a function.

5 0
4 years ago
Determine the axis of symmetry of a parabola whose graph is given below. Give your answer in the form X equals H
sergey [27]

Answer:

x = -3

Step-by-step explanation:

3 0
3 years ago
Which linear inequality is represented by the graph?
larisa [96]

Answer:

y>3x-4

Step-by-step explanation:

8 0
3 years ago
Suppose the time a child spends waiting at for the bus as a school bus stop is exponentially distributed with mean 7 minutes. De
Gala2k [10]

Answer:

The probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.

Step-by-step explanation:

Let the random variable <em>X</em> represent the time a child spends waiting at for the bus as a school bus stop.

The random variable <em>X</em> is exponentially distributed with mean 7 minutes.

Then the parameter of the distribution is,\lambda=\frac{1}{\mu}=\frac{1}{7}.

The probability density function of <em>X</em> is:

f_{X}(x)=\lambda\cdot e^{-\lambda x};\ x>0,\ \lambda>0

Compute the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning as follows:

P(6\leq X\leq 9)=\int\limits^{9}_{6} {\lambda\cdot e^{-\lambda x}} \, dx

                      =\int\limits^{9}_{6} {\frac{1}{7}\cdot e^{-\frac{1}{7} \cdot x}} \, dx \\\\=\frac{1}{7}\cdot \int\limits^{9}_{6} {e^{-\frac{1}{7} \cdot x}} \, dx \\\\=[-e^{-\frac{1}{7} \cdot x}]^{9}_{6}\\\\=e^{-\frac{1}{7} \cdot 6}-e^{-\frac{1}{7} \cdot 9}\\\\=0.424373-0.276453\\\\=0.14792\\\\\approx 0.148

Thus, the probability that the child must wait between 6 and 9 minutes on the bus stop on a given morning is 0.148.

6 0
3 years ago
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