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3 years ago

(x+2)-(X-2) PLEASE HELP ME

Mathematics

2 answers:

guajiro [1.7K]3 years ago

8 0

Answer:

4 is the answer

Step-by-step explanation:

x+2-x+2= 4

lilavasa [31]3 years ago

6 0

Answer:

If you are talking about subtracting it is 4

Step-by-step explanation:

(x+2)-(X-2)=4

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D. 30%?

I'm not 100% sure but I think its D. Sorry if it's wrong.

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Does y=3^t represent growth or decay?

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Decay

Step-by-step explanation:

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A photoconductor film is manufactured at a nominal thickness of 25 mils. The product engineer wishes to increase the mean speed

AURORKA [14]

Answer:

A 98% confidence interval estimate for the difference in mean speed of the films is [-0.042, 0.222].

Step-by-step explanation:

We are given that Eight samples of each film thickness are manufactured in a pilot production process, and the film speed (in microjoules per square inch) is measured.

For the 25-mil film, the sample data result is: Mean Standard deviation 1.15 0.11 and For the 20-mil film the data yield: Mean Standard deviation 1.06 0.09.

Firstly, the pivotal quantity for finding the confidence interval for the difference in population mean is given by;

P.Q. = $\frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t__n_1_+_n_2_-_2$

where, $\bar X_1$ = sample mean speed for the 25-mil film = 1.15

$\bar X_1$ = sample mean speed for the 20-mil film = 1.06

$s_1$ = sample standard deviation for the 25-mil film = 0.11

$s_2$ = sample standard deviation for the 20-mil film = 0.09

$n_1$ = sample of 25-mil film = 8

$n_2$ = sample of 20-mil film = 8

$\mu_1$ = population mean speed for the 25-mil film

$\mu_2$ = population mean speed for the 20-mil film

Also, $s_p =\sqrt{\frac{(n_1-1)s_1^{2}+ (n_2-1)s_2^{2}}{n_1+n_2-2} }$ = $\sqrt{\frac{(8-1)\times 0.11^{2}+ (8-1)\times 0.09^{2}}{8+8-2} }$ = 0.1005

Here for constructing a 98% confidence interval we have used a Two-sample t-test statistics because we don't know about population standard deviations.

So, 98% confidence interval for the difference in population means, ( $\mu_1-\mu_2$ ) is;

P(-2.624 < $t_1_4$ < 2.624) = 0.98 {As the critical value of t at 14 degrees of

freedom are -2.624 & 2.624 with P = 1%}

P(-2.624 < $\frac{(\bar X_1 -\bar X_2)-(\mu_1- \mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < 2.624) = 0.98

P( $-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < $2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < ) = 0.98

P( $(\bar X_1-\bar X_2)-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ) = 0.98

98% confidence interval for ( $\mu_1-\mu_2$ ) = [ $(\bar X_1-\bar X_2)-2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+2.624 \times {s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ]

= [ $(1.15-1.06)-2.624 \times {0.1005 \times \sqrt{\frac{1}{8}+\frac{1}{8} } }$ , $(1.15-1.06)+2.624 \times {0.1005 \times \sqrt{\frac{1}{8}+\frac{1}{8} } }$ ]

= [-0.042, 0.222]

Therefore, a 98% confidence interval estimate for the difference in mean speed of the films is [-0.042, 0.222].

Since the above interval contains 0; this means that decreasing the thickness of the film doesn't increase the speed of the film.

7 0

3 years ago

A bottle contains 255 coins. 1/3 of the coins are ?1.00. 110 of the coins are 50p coins. the rest are 20p coins. work out the to

aliina [53]

Answer:

£152.

Step-by-step explanation:

We have been given that a bottle contains 255 coins. 1/3 of the coins are £1.00.

Let us find 1/3 of 255 to find the number of £1 coins.

$\£1\text{ coins}=\frac{1}{3}\times 255$

$\£1\text{ coins}=85$

This means we have £85.

We are also told that 110 of the coins are 50 p coins.

$\text{Value of 50 p coins}=\£0.50\times 110$

$\text{Value of 50 p coins}=\£55$

Let us figure out number of 20 p coins by subtracting the number of £1 coins and 50 p coins from 255.

$\text{Number of 20 p coins}=255-(85+110)$

$\text{Number of 20 p coins}=255-(195)$

$\text{Number of 20 p coins}=60$

$\text{Value of 20 p coins}=\£0.20\times 60$

$\text{Value of 20 p coins}=\£12$

Now let us find total value of the coins contained in the bottle by adding the values of £1 coins, 50 p coins and 20 p coins.

$\text{The total value of the coins}=\£85+\£55+\£12$

$\text{The total value of the coins}=\£152$

Therefore, the total value of the coins contained in the bottle is £152.

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Answer:

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