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3 years ago

I'm really bad at math!

Mathematics

2 answers:

lidiya [134]3 years ago

6 0

Answer:

i think its 5b + 45

Step-by-step explanation:

im not 100% sure what your asking for but i just distributed the -3 to the 8 and -26 and got 5b -9 -24 + 78. then i combined the like terms to get 5b + 45.

hope this helps !!

Igoryamba3 years ago

3 0

Given : (5b - 9) - 3(8 - 26)

⇒ (5b - 9) - 3(-18)

⇒ (5b - 9) + 54

⇒ 5b - 9 + 54

⇒ 5b + 45

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The gas tank on the back of a tanker truck can be equated to a cylinder with a diameter of 8 feet and a length of 19 feet. A gal

Leokris [45]

9514 1404 393

Answer:

driver's tank: 30,427 lb
farmer's tank: 12,811 lb

Step-by-step explanation:

The formula for the volume of a cylinder is ...

V = πr^2·h . . . radius r, height h

The radius of the driver's tank is half its diameter, so is (8 ft)/2 = 4 ft. Then the volume of that tank is ...

V = π(4 ft)^2·(19 ft) = 304π ft^3

Each cubic foot of gasoline has a mass of ...

(1728 in^3/ft^3)(0.0262 lb/in^3) = 45.2736 lb/ft^3

Then the total mass in the driver's full tank is ...

(304π ft^3)(45.2736 lb/ft^3) ≈ 43,238.3 lb

The farmer's tank is a scaled-down version of the driver's tank. It's volume will be scaled by the cube of the linear scale factor, so will be (2/3)^3 = 8/27 of the volume of the driver's tank.

The farmer's tank will hold a mass of (43,238.3 lb)(8/27) ≈ 12,811 lb.

The amount remaining in the driver's tank is 43,238 -12,811 = 30,427 lb.

6 0

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

3 years ago

6b. Create a table that shows the relationship between the amount of

goblinko [34]

Answer:

120 x W = M

Step-by-step explanation:

If she makes 120 per week working then the number of the week she has work multiplied by 120 is her total for whichever week

8 0

3 years ago

What is the gcf of 28 and 60

Vikki [24]

Answer:

Step-by-step explanation:

We need to find the factors of each number

28 -

1x28

2x14

4x7

60 -

1x60

2x30

3x20

4x15

5x12

6x10

then we need to find the largest number they have in common

in this case it is 4

3 0

3 years ago

I have no idea how to do this, can someone please help?

Sveta_85 [38]

Assume your line starts at zero, your first point is (-3,5) meaning your have a slope of 5/-3
[ f(x) = 5/-3x + b]

4 0

3 years ago