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2 years ago

Can someone help solve both of these? ASAP

Mathematics

1 answer:

zhuklara [117]2 years ago

7 0

Question 4

$\sin 30^{\circ}=\frac{y}{14\sqrt{3}}\\\\\frac{1}{2}=\frac{y}{14\sqrt{3}}\\\\\boxed{y=7\sqrt{3}}\\\\\\\\\cos 30^{\circ}=\frac{x}{14\sqrt{3}}\\\\\frac{\sqrt{3}}{2}=\frac{x}{14\sqrt{3}}\\\\\boxed{x=21}$

Question 5

$\cos 45^{\circ}=\frac{x}{28}\\\\\frac{1}{\sqrt{2}}=\frac{x}{28}\\\\x=\frac{28}{\sqrt{2}}\\\\\boxed{x=14\sqrt{2}}\\\\\\\\\sin 45^{\circ}=\frac{y}{28}\\\\\frac{1}{\sqrt{2}}=\frac{y}{28}\\\\y=\frac{28}{\sqrt{2}}\\\\\boxed{y=14\sqrt{2}}$

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A farmer has a rectangle field with a perimeter of 5.4 miles and a length of 2.3 miles. Find the width of the field. The width o

Alja [10]

Answer:

Step-by-step explanation:

perimeter = 2*(length + width)

53 m = 2*(2.3 m +w) . . . . an equation to find the width, w

26.5 m = 2.3 m +w . . . . . divide by 2

24.2 m = w . . . . . . . . . . . subtract 2.3 m

The width of the rectangle is 24.2 m

hope it helps

6 0

3 years ago

Read 2 more answers

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

3 years ago

I played 5 hours of video games over a 3 day weekend, and 60 hours of video games in the month

den301095 [7]

Answer:

The ratios are not equivalent

Step-by-step explanation:

First, in order to find the ratios, we have to change the month to the amount of days.

There are 30 days in April, so we will use this for the ratio.

The ratio for the 3 day weekend is 5:3

The ratio for the month of April is 60:30, which can be simplified to 2:1

So, the ratios are not equivalent, because 5:3 is not equal to 2:1

8 0

3 years ago

I need the answer with work

Alenkasestr [34]

1. Regroup terms
x^3 - 6 * 10x^2x-8/ x-3
2. Use product rule
x^3 - 6 * 10x^ 2+1 - 8/ x-3
3. Simplify
x^3 - 6 * 10 x^3 - 8/ x-3
4. Simplify further
x^3 - 60x^3 - 8/ x-3
5. Simplify the last time
-59x^3-8/x-3

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3 years ago

Find the absolute value of... |7| -7 or 7?

erastova [34]

The answer to this is 7

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3 years ago

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