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sasho [114]
3 years ago
14

A + 3(a - 1)= 3(2 +1)

Mathematics
1 answer:
N76 [4]3 years ago
5 0
A+3a-3=6+3. 4a=12. a=3.
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Find 16% of 370. Round to the nearest 10th if necessary.
svetoff [14.1K]
Multiply 370 by 0.16 get get 16%

370 x 0.16 = 59.2%
8 0
3 years ago
Which statements are true about the graph of the function f(x) = 6x – 4 + x2?
kaheart [24]
Write the function as
f(x) = x² + 6x - 4

Note that
x² + 6x = (x + 3)² - 9

Therefore
f(x) = (x  3)² - 9 - 4
      = (x + 3)² - 13

The vertex is at (-3, -13)., and the curve opens upward.
The axis of symmetry is x = -3.

Answer:
The vertex of the function is (-3, 13).
The graph increases over the interval (-3, -∞)


3 0
3 years ago
Write an expression equivalent to (7x + 5) - (3x - 5)​
KiRa [710]

Answer:

= 2(2x + 5)

Step-by-step explanation:

Given that:

= (7x + 5) - (3x - 5)​

By simplifying

"-" sign before bracket will alter the inner signs

= 7x + 5 - 3x + 5

Adding like terms

= 4x + 10

By taking 2 common we get:

= 2(2x + 5)

i hope it will help you!

6 0
3 years ago
Study the Proof Given.
saul85 [17]
2: Midpoint of segment AC is given as B3: Given4: Knowing that AC~=BC, AC~= DE
7 0
3 years ago
A consumer group has determined that the distribution of life spans for gas ranges (stoves) has a mean of 15.0 years and a stand
Art [367]

Answer:

b. Mean = 1.6 years, standard deviation - 0.92 years, shape: approximately Normal.

Step-by-step explanation:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Subtraction of normal Variables:

When we subtract normal variables, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.

A consumer group has determined that the distribution of life spans for gas ranges (stoves) has a mean of 15.0 years and a standard deviation of 4.2 years. Sample of 35:

This means that:

\mu_G = 15

s_G = \frac{4.2}{\sqrt{35}} = 0.71

The distribution of life spans for electric ranges has a mean of 13.4 years and a standard deviation of 3.7 years. Sample of 40:

This means that:

\mu_E = 13.4

s_E = \frac{3.7}{\sqrt{40}} = 0.585

Which of the following best describes the sampling distribution of the difference in mean life span of gas ranges and electric ranges?

Shape is approximately normal.

Mean:

\mu = \mu_G - \mu_E = 15 - 13.4 = 1.6

Standard deviation:

s = \sqrt{s_G^2+s_E^2} = \sqrt{0.71^2+0.585^2} = 0.92

So the correct answer is given by option b.

8 0
3 years ago
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