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2 years ago

How do u upload a pict of a problem

Mathematics

2 answers:

Trava [24]2 years ago

5 0

Answer:

take a screenshot, then copy and paste it to your question

Step-by-step explanation:

Zinaida [17]2 years ago

3 0

Answer:

When you click ask your question, find the paper clip icon which is a way to attach a photo. Click the paper clip and select the photo or file that you want to upload.

Step-by-step explanation:

You might be interested in

Is 0.7 =0.07 reasonable? explain why if its not or it is

MrRissso [65]

Answer:

Step-by-step explanation:

No, because 0.7 is greater than 0.07 {Tip: Look for place of the decimal}

0.7 is written as 7/10 and 0.07 is written as 7/100

0.7 which is 7/10 is also written as 70/100 Hence, it clearly that 70/100 is greater than 7/100

0.7 is greater than 0.07 because to compare between two decimal numbers, you should look at the number after the decimal point and 7 is greater than 0 in the tenth place.

[RevyBreeze]

5 0

2 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

Use the continuous compound interest formula to find the indicated value. A=90,000; P=65,452; r=9.1%; t=? t=years (Do not round

Inga [223]

Using continuous compounding, we have:
90000=65452 x e^.091t
1.3750534743017784024934303000672=e^.091t
ln 1.3750534743017784024934303000672=ln e^.091t=.091t ln e=0.091t
t=3.5 years

3 0

3 years ago

A square pyramid has a base length of 4 inches. The height of each triangular face is 12 inches. What is the surface area of the

Wittaler [7]

Answer:

The surface area of the given pyramid is 112 $inches^{2}$ .

Step-by-step explanation:

Base length of the pyramid = 4 inches

Area of the base = $length^{2}$

= $4^{2}$

= 16

Area of its base = 16 $inches^{2}$

Area of one of its triangular surface = $\frac{1}{2}$ x base x height

= $\frac{1}{2}$ x 4 x 12

= 2 x 12

= 24 $inches^{2}$

Area of all its four triangular surfaces = 4 x 24

= 96 $inches^{2}$

surface area of the pyramid = sum of areas of all its surfaces

= 16 + 96

= 112 $inches^{2}$

The surface area of the given pyramid is 112 $inches^{2}$ .

8 0

2 years ago

Solve the following system of equations. If there is no solution, write DNE in each coordinate of the

Rainbow [258]

Answer:

1) solution is y = -7

2.) DNE

3.) Solution is y = -2

Step-by-step explanation:

1.) 2+3 = y + 12

Add the LHS and make y the subject of formula

5 = y + 12

Y = 5 - 12

Y = - 7

The solution of the equation is - 7

2.) 2 + 13 = 1 +8

Since there is no unknown variable and the sum of the numbers in the left hand side ( LHS ) is not equal to the sum of the numbers in the right hand side ( RHS ), it will be concluded that there is no solution in the equation.

3.) y - 7 = 2 - 11

Sum the RHS and make y the subject of formula

Y - 7 = -9

Y = -9 + 7

Y = -2

The solution of the equation is -2

6 0

2 years ago