(5i^{2} -3) * (6i *2^{6})

Soup A has 5 grams of sodium in every 12 cups. Soup B has 8 grams of sodium in every 13 cups. Which soup has more sodium per cup

andrew11 [14]

Answer:

Soup B has more sodium per cup.

Step-by-step explanation:

If we're doing sodium per cup, then we have to divide 5 by 12, and 8 by 13.

5÷12=~0.4 grams

8÷13=~0.6 grams

0.6>0.4

So B has more sodium per cup.

6 0

2 years ago

The number of contaminating particles on a silicon wafer prior to a certain rinsing process was determined for each wafer in a s

GalinKa [24]

Answer:

1. a, 99% b71%

2. 64%, 44%

3. Number of particles 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14

frequencies: 1, 2, 3, 12, 11, 15, 18, 10, 12, 4, 5, 3, 1, 2, 1

Relative frequencies 0.01, 0.02,0.03,0.12,0.11,0.15, 0.18,0.1, 0.12,0.04, 0.05,0.03,0.01,0.02,0.01

Step-by-step explanation:

Number of particles: 0, 1, 2, 3, 4, , 5, 6, 7

Frequency: 1, 2, 3, 12, 11, 15, 18, 10

Number of particles: 8, 9, 10, 11, 12, 13, 14

Frequency: 12, 4, 5, 3, 1, 2, 1

we solve by saying that . one of th smaples wafer is with any particles. we get the proportion by subtracting 1 from the total divided by 100

100-1/100

.99

at least five particles , then we can subtract the number of those with particles less than five from 100 , divided by 100

100-(11+12+3+2+1)=71/100

0.71 or 71%

b. those particles within 5 and 10 inclusive

15+18+10+12+4+5=64/100

0.64

0r 64%

five and 10 exclusively will be

18+10+12+4=44/100

0.44

44%

c. Draw an histogram. first we have frequency table first

Number of particles 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14

frequencies: 1, 2, 3, 12, 11, 15, 18, 10, 12, 4, 5, 3, 1, 2, 1

Relative frequencies 0.01, 0.02,0.03,0.12,0.11,0.15, 0.18,0.1, 0.12,0.04, 0.05,0.03,0.01,0.02,0.01

Relative frequencies is derived by dividing each frequencies over the total frequencies

wecan say that it is unimodal or say that it is slightly positively skewed

4 0

3 years ago

Two angle are supplementary angle A has a measure of 3x+12 and angle B has a measure of 4x+28, what is the measure of each angle

RideAnS [48]

Answer:

A measures 72 and B measures 108

Step-by-step explanation:

If the angles are supplementary, that means that they add up to equal 180. 3x + 12 + 4x + 28 = 180. Doing some algebra there gives you that 7x = 140 and x = 20. Sub in 20 to angle A to get 3(20) + 12 = 72, and sub in 20 to angle B to get 4(20) + 28 = 108. And of course, 108 + 72 = 180.

5 0

3 years ago

A simulation was conducted using 10 fair six-sided dice, where the faces were numbered 1 through 6. respectively. All 10 dice we

kompoz [17]

Answer:

C) a sample distribution of a sample mean with n = 10

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

Step-by-step explanation:

Here, the random experiment is rolling 10, 6 faced (with faces numbered from 1 to 6) fair dice and recording the average of the numbers which comes up and the experiment is repeated 20 times.So, here sample size, n = 20 .

Let,

$X_{ij}$ = The number which comes up on the ith die on the jth trial.

∀ i = 1(1)10 and j = 1(1)20

Then,

$E(X_{ij})$ = $\frac {1 + 2 + 3 + 4 + 5 + 6}{6}$

= 3.5 ∀ i = 1(1)10 and j = 1(1)20

and,

$E(X^{2}_{ij}$ = $\frac {1^{2} + 2^{2} + 3^{2} + 4^{2} + 5^{2} + 6^{2}}{6}$

= $\frac {1 + 4 + 9 + 16 + 25 + 36}{6}$

= $\frac {91}{6}$

$\simeq$ 15.166667

so, $Var(X_{ij}$ = $(E(X^{2}_{ij} - {(E(X_{ij})}^{2})$

$\simeq 15.166667 - 3.5^{2}$

= 2.91667

and $\sigma_{X_{ij}}$ = $\sqrt {2.91667}[/tex [tex]\simeq 1.7078261036$

Now we get that,

$Y_{j} = \frac {\sum_{j = 1}^{20}X_{ij}}{20}$

We get that $Y_{j}'s$ are iid RV's ∀ j = 1(1)20

Let, ${\overline}{Y} = \frac {\sum_{j = 1}^{20}Y_{j}}{20}$

So, we get that $E({\overline}{Y}) = E(Y_{j})$

= $E(X_{ij}$ for any i = 1(1)10

= 3.5

and,

$\sigma_{({\overline}{Y})} = \frac {\sigma_{Y_{j}}}{\sqrt {20}} = \frac {\sigma_{X_{ij}}}{\sqrt {20}} = \frac {1.7078261036}{\sqrt {20}} [tex]\simeq 0.38$

Hence, the option which best describes the distribution being simulated is given by,

C) a sample distribution of a sample mean with n = 10

$\mu_{{\overline}{X}} = 3.5$

and $\sigma_{{\overline}{Y}} = 0.38$

6 0

2 years ago

A kite is stuck in a 36-ft tree. If the angle of elevation from the kite and the ground is 12, find the length of the kite.

yanalaym [24]

Answer:

173.15 ft.

Step-by-step explanation:

Please find the attachment.

Let L be the length of kite.

We have been given that a kite is stuck in 36-ft tree. The angle of elevation from the kite and the ground is 12.

We can see from our attachment that kite and tree form a right triangle. height of tree is opposite of angle of elevation and length of kite is hypotenuse.

Since we know that Sine relates the opposite and hypotenuse of a right triangle, so we will use sine to find the length of kite.

$sin(12)=\frac{36}{L}$

$L=\frac{36}{sin(12)}$

$L=\frac{36}{0.207911690818}$

$L=173.1504364105882792\approx 173.15$

Therefore, the length of kite is 173.15 ft.

4 0

2 years ago