All categories

3 years ago

I WILL GIVE CROWN PLS HELP NEED RIGHT ANSWER

Mathematics

2 answers:

FromTheMoon [43]3 years ago

4 0

Answer:

Step-by-step explanation:

Surely you can figure out how many stairs she has to climb. She has to go down 4, back up to 0, and then up another 4, then back to 0

aev [14]3 years ago

3 0

Answer: 21.4 ft

Step-by-step explanation: considering all levels are same size, ÷ depth of 7th, the deepest level with number of levels. 149.8 ÷ 7 = 21.4

You might be interested in

Michell bought 5 packs of crayons for $13.75 what's the coust of one pack?

Keith_Richards [23]

All you would do is divide for this problem so,

13.75 ÷ 5 = 2.75

So one pack would cost 2.75

(:

5 0

3 years ago

Read 2 more answers

Solve problem-4/9 - 8/9 =?

mrs_skeptik [129]

it is the same denominator therefore only the numerators are added and the sign is maintained. The answer is -12/9 but it can be reduced to -4/3.

4 0

3 years ago

Read 2 more answers

A gumball machine has various colors of gumballs. There is a 10% chance that the next

iogann1982 [59]

Answer:

It is red

Step-by-step explanation:

4 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

2 years ago

Tran is solving the quadratic equation 2x2 – 4x – 3 = 0 by completing the square. His first four steps are shown in the table.

Nady [450]

The following steps of solving for the roots of 2x² - 4x -3 = 0 were retrieved from another source Step 1 2x² - 4x = 3 Step 2 2(x² - 2x) = 3 Step 3 2(x² - 2x + 1) = 3 + 1 Step 4 2(x - 1)² = 4 From this, we can see that on Step 3, Tran made a mistake of adding 1 to 3. As we can see, 2(x² - 2x + 1) = 2x² - 4x + 2. That means, instead of adding 1, it should have been 2. Therefore, the step that Tran first made an error is Step 3.

3 0

3 years ago

Read 2 more answers