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3 years ago

Help me with #4

Mathematics

1 answer:

Sladkaya [172]3 years ago

6 0

Step-by-step explanation:

a. lim(x→2) [g(x) + h(x)]

Use additive property of limits.

= lim(x→2) g(x) + lim(x→2) h(x)

= 0 + 5

= 5

b. lim(x→2) [3 h(x)]

Use multiplication property of limits.

= [lim(x→2) 3] [lim(x→2) h(x)]

= 3 lim(x→2) h(x)

= 3 (5)

= 15

c. lim(x→2) [g(x) h(x)]

Use multiplication property of limits.

= [lim(x→2) g(x)] [lim(x→2) h(x)]

= (0) (5)

= 0

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2564.0 millimeters tall. If the boards are too long they must be trimmed, and if the boards are too short they cannot be used. A

Leokris [45]

Answer:

We conclude that the board's length is equal to 2564.0 millimeters.

Step-by-step explanation:

We are given that a sample of 26 is made, and it is found that they have a mean of 2559.5 millimeters with a standard deviation of 15.0.

Let $\mu$ = population mean length of the board.

So, Null Hypothesis, $H_0$ : $\mu$ = 2564.0 millimeters {means that the board's length is equal to 2564.0 millimeters}

Alternate Hypothesis, $H_A$ : $\mu$ $\neq$ 2564.0 millimeters {means that the boards are either too long or too short}

The test statistics that will be used here is One-sample t-test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s }{\sqrt{n}} }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean length of boards = 2559.5 millimeters

s = sample standard deviation = 15.0 millimeters

n = sample of boards = 26

So, the test statistics = $\frac{2559.5-2564.0}{\frac{15.0 }{\sqrt{26}} }$ ~ $t_2_5$

= -1.529

The value of t-test statistics is -1.529.

Now, at a 0.05 level of significance, the t table gives a critical value of -2.06 and 2.06 at 25 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the board's length is equal to 2564.0 millimeters.

4 0

4 years ago

To find the product of 42.12 and 10³, move the decimal point in 42.12 __ places to the right because 10³ has __ zeros.

Yakvenalex [24]

<h3>To find the product of 42.12 and 10^3, move the decimal point in 42.12 3 places to the right because 10^3 has 3 zeros</h3>

Solution:

Given that,

$\text{ product of } 42.12 \text{ and } 10^3$

Which means,

$42.12 \times 10^3$

Here, the exponent of 10 is positive ( which is 3)

When the exponent is positive, we have to move the decimal point to right

When you multiply a number by a power of 10, ( 10!, 10^2, and so on ) move the decimal point of the number to the right the same number of places as the number of zeros in the power of 10

Here, exponent is 3 , therefore move the decimal point right 3 places in 42.12

Therefore,

$42.12 \times 10^3 = 42120$