Mr cold war is coming back......whats 4 to the 7th power ×16 to the 3rd power?

Ratio..A candy store sells snacks by weight. A six ounce bag of candy cost $3.60. Predict the cost of a 2ounce bag.

KIM [24]

Answer:

A 2 ounce bag costs $1.2.

Step-by-step explanation:

With the information provided, you can use a role of three to find the cost of a 2 ounce bag given that you know that a six ounce bag costs $3.60.

6 ounce bag → 3.60

2 ounce bag → x

x=(2*3.60)/6

x=7.20/6

x=1.2

According to this, the answer is that a 2 ounce bag costs $1.2.

7 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

A movie cost $3,254,107 to produce. What digit is in the hundred thousands place?

Alla [95]

The two in 3,254,107 is in the hundred thousands place

3 0

3 years ago

How do I solve this?

ozzi

The answer is A) √3
cot(-5π/6)=1/tan(-5π/6)
=1/tan(-(5π/6)
Use the property: tan(-x)=-tan(x)
=1/tan(5π/6)
-(1/tan(5π/6))
tan(5π/6)=-√3/3
-(1/-√3/3)
-(1/-√3/3)
=√3

4 0

2 years ago

Read 2 more answers

HELP i need the answers to these! explain if possible:)

IrinaVladis [17]

so one: the middle Angles are all 90 degrees triangles equal 180, the angle opposite of "x" is equal to 34 so 180-90-34=x

two: both angles are equal set it up like an equation and solve for n

three: the corner angle of the right triangle is equal to 61 as well again do 180-(61×2)=x

4: do this equation (<ktu)2+ x = 180 then solve for x

5: not exactly.sure what its asking

hope this helps

5 0

3 years ago