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3 years ago

Solve the inequality. 3x + 7 > 4x + 2

Mathematics

2 answers:

ankoles [38]3 years ago

7 0

5>x because you can subtract 5 from both sides and 3x from both sides

Nutka1998 [239]3 years ago

5 0

By solving this kind inequality, we would then have to find what does (x) equal, and then from here, we would determine which would be greater.

$rearrange: \\ 3*x+7-(4*x+2)\ \textgreater \ 0 \\ \\ Multiply \ both \ sides \ by \ (-1) \\ \\$

Then after, we add both sides by 5.

3x + 7= $3*5+7$ =22

4x + 2= $4*5+2$ =22

3x + 7 (x) 4x + 2

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xeze [42]

Answer:

3rd one

Step-by-step explanation:

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3 years ago

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

2 years ago

Find the volume of a half sphere with a radius of 10

Ugo [173]

The volume of a sphere is calculated like so:
v = 4*pi*r^3/3
so half sphere will have half that volume:
v = (4*pi*r^3/3)(1/2)
v = 4*pi*r^3/6
plug in the data:
v = 4*pi*(10)^3/6
v = 2094.4
that is the volume of the half sphere

7 0

3 years ago

there were 7 songs on one side of a record. it took 20 minutes and 18 seconds to play the side what is the average number of sec

kompoz [17]

Answer:

174 seconds

Step-by-step explanation:

I tried my best please don't report

7 0

2 years ago

At t=0, a rock is dropped from rest from the top of a building 256 ft high. With what velocity will it strike the ground? What i

Rasek [7]

Answer:

128 ft / second.

Step-by-step explanation:

It's acceleration is due to gravity and is 32 ft s^-2.

To find the velocity when it hits the ground we use the equation of motion

v^2 = u^2 + 2gs where u = initial velocity, g = acceleration , s = distance.

v^2 = 0^2 + 2 * 32 * 256

v^2 = 16384

v = 128 ft s^-1.

4 0

3 years ago