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2 years ago

Iam not sure about this questions answer can u give answer Q10,11,12

Mathematics

1 answer:

irga5000 [103]2 years ago

7 0

Answer:

10. First you multiply both sides by 3, making it 2x + 5 = 24

Then you subtract 5 from both sides making it 2x = 19

You dive both sides by 2 making x = 9.5

The answer is x = 9.5

11. First you need to find how much you need to pay each year. 12,000 x 0.05

you can make it easier by changing it to 120 x 5 = 600

You need to pay that for three years so 600 x 3 = 1800

Plus you need to pay the original money. 12,000 + 1,800 = 13,800

The answer is $13,800

12. First you change kilometers into centimeters. 172 km is 172,000 meters. 172,000 meters is 17,200,000

Then, you need to do 17,200,000 ÷ 500,000 but you can make it simple and do 172 ÷ 5 = 34.4

The answer is 34.4 cm

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5 0

2 years ago

Read 2 more answers

Place parentheses in each equation, if needed, to make each equation true. WILL GIVE BRAINLIEST AND FRIEND YOU. PLEASE ANSWER VE

bezimeni [28]

A. 7 + 3 x (2 + 4) = 25
7 + 3 x 6 = 25
7 + 18 = 25
25 = 25
B. 8^2 divided by 4 x 8 = 2
64 divided by 4 x 8 = 2
64/32 = 2
2 = 2
C. (16 + 8) - 5 x 2 = 14
24 - 5 x 2 = 14
24 - 10 = 14
14 = 14
For each problem use the basis of the acronym PEMDAS.
P arentheses (grouping symbols)
E xponents
M ultiplication
D ivision
A ddition
S ubtraction

Evaluate each step in that order.

3 0

3 years ago

What are the vertices of A'B'C if ABC is dilated by a scale factor of 2?

malfutka [58]

The vertices of A'B'C' if A(0, 10), B(8, 6), C(4, 2) is dilated by a scale factor of 2 are;

C. A'(0, 20), B'(16, 12), C'(8, 4)

<h3>Which method can be used to find the vertices of A'B'C'?</h3>

Given that triangle ABC is dilated by a scale factor of 2, we have;

A'B' = 2 × AB

A'C' = 2 × AC

B'C' = 2 × BC

Length of AB = √((8-0)^2 + (6-10)^2) = 4•√5

Length of AC = √((4-0)^2 + (2-10)^2) = 4•√5

Length of BC = √((8-4)^2 + (6-2)^2) = 4•√2

By multiplying the given coordinates by the scale factor, we have;

A' = 2 × (0, 10) = (0, 20)

B' = 2 × (8, 6) = (16, 12)

C' = 2 × (4, 2) = (8, 4)

A'B' = √((16-0)^2 + (12-20)^2) = 8•√5

A'C' = √((8-0)^2 + (4-20)^2) = 8•√5

B'C' = √((16-8)^2 + (12-4)^2) = 8•√2

Therefore when we have;

A'(0, 20), B'(16, 12), C'(8, 4), we get;

A'B' = 2 × AB

A'C' = 2 × AC
B'C' = 2 × BC

The correct option is therefore;

C. A'(0, 20), B'(16, 12), C'(8, 4)

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5 0

1 year 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

<em>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.</em>

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

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

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

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

= [-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

A machine fills containers with 35 ounces of raisins. After the containers are filled, another machine weighs them. If the conta

EleoNora [17]

Answer:

a think only thing thats close to my data

Step-by-step explanation:

7 0

2 years ago