NEED HELP ASAP trig question!! Need to find y!!

Answer:

see below

Step-by-step explanation:

Difference is subtract

(8-2)

Then add this to x

(8-2) +x

6+x

False.

2x10^6 you move the decimal point 6 places to the right. ( add 6 zeros after the 2)

2x 10^6 = 2,000,000

The value of y will be equal to 4. The correct option is D.

Expression in maths is defined as the collection of the numbers variables and functions by using signs like addition, subtraction, multiplication, and division.

Numbers (constants), variables, operations, functions, brackets, punctuation, and grouping can all be represented by mathematical symbols, which can also be used to indicate the logical syntax's order of operations and other features.

The given expression 8(y + 2) = 48 will be solved for y as below:-

8(y + 2) = 48

Divide both sides by 8 and solve.

[ 8 (y + 2) ] / 8 = 48 / 8

y + 2 = 6

Substract 2 from both the sides to get the value of y.

y + 2 - 2 = 6 -2

y = 4

Therefore, the value of y will be equal to 4. The correct option is D.

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Answer:

Step-by-step explanation:

Given that:

Mean = 3.34

sample size = 26

sample mean = 2.7

standard deviation = 1.17

level of significance = 0.10

The null hypothesis and the alternative hypothesis can be computed as follows:

[tex]\mathtt{H_o: \mu \geq 3.34} \\ \\ \mathtt{H_1: \mu < 3.34}[/tex]

degree of freedom = n - 1

degree of freedom = 26 -1

degree of freedom =  25

level of significance = 0.10

Since the alternative hypothesis contains <, then the test is left tailed

[tex]\mathtt{t_{\alpha, df} = t_{0.10, 25}}[/tex]

[tex]\mathtt{t_{0.10, 25}}[/tex] = - 1.316

The rejection region therefore consist of all values smaller than - 1.316, therefore ; reject [tex]H_o[/tex] if t < -1.316

The test statistics can be computed as follows:

[tex]t = \dfrac{X - \mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

[tex]t = \dfrac{2.7 - 3.34}{\dfrac{1.17}{\sqrt{26}}}[/tex]

[tex]t = \dfrac{-0.64}{\dfrac{1.17}{5.099}}[/tex]

t = - 2.789

Decision Rule:  To reject the null hypothesis if the t test lies in the rejection region or less than the rejection region.

Conclusion: We  reject the null hypothesis since t = (- 2.789) < -1.316. Then we conclude that  the mean number of residents in the retirement community household is less than 3.34 persons.

Answer:

[tex]\Large \boxed{\sf \bf \ \ y=\dfrac{1}{16}(a-3)^2-2 \ \ }[/tex]

Step-by-step explanation:

Hello, please consider the following.

When the parabola equation is like

[tex]y=a(x-h)^2+k[/tex]

The vertex is the point (h,k) and the focus is the point (h, k+1/(4a))

As the vertex is (3,-2) we can say that h = 3 and k = -2.

We need to find a.

The focus  is (3,2) so we can say.

[tex]2=-2+\dfrac{1}{4a}\\\\\text{*** We add 2. ***}\\\\\dfrac{1}{4a}=2+2=4\\\\\text{*** We multiply by 4a. ***}\\\\16a=1\\\\\text{*** We divide by 16. ***}\\\\a=\dfrac{1}{16}[/tex]

So an equation for the parabola is.

[tex]\large \boxed{\sf y=\dfrac{1}{16}(a-3)^2-2 }[/tex]

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

Answer:

- At any time t, the population is:

P = 375t² + 3000t + 1000

- At time t = 3 days, the population is:

P = 13,375

Step-by-step explanation:

Given the rate of change of the population of bacteria as:

dP/dt = 3000/(1 + 0.25t)

we need to rewrite the given differential equation, and solve.

Rewriting, we have:

dP/3000 = (1 + 0.25t)dt

Integrating both sides, we have

P/3000 = t + (0.25/2)t² + C

P/3000 = t + 0.125t² + C

When t = 0, P = 1000

So,

1000/3000 = C

C = 1/3

Therefore, at any time t, the population is:

P/3000 = 0.125t² + t + 1/3

P = 375t² + 3000t + 1000

At time t = 3 days, the population is :

P = 375(3²) + 3000(3) + 1000

= 3375 + 9000 + 1000

P = 13,375

Answer:

The percentage increase in the production cost of the printer is 3%.

Step-by-step explanation:

We are given that the production of a printer consists of the cost of raw material at 100 dollars the cost of overheads at 80$ and wages at 120$.

Also, the cost of raw materials and overheads are increased by 11% and 20% respectively while wages are decreased by 15%.

Cost of raw material = $100

Cost of overheads = $80

Cost of wages = $120

So, the total cost of the printer = $100 + $80 + $120

                                                   = $300

Now, the increase in the cost of raw material = $100 + 11% of $100

                                                                           = [tex]\$100 + (\frac{11}{100} \times \$100)[/tex]

                                                                           = $100 + $11 = $111

The increase in the cost of overheads = $80 + 20% of $80

                                                                = [tex]\$80 + (\frac{20}{100} \times \$80)[/tex]

                                                                = $80 + $16 = $96

The decrease in the cost of wages = $120 - 15% of $120

                                                          = [tex]\$120 - (\frac{15}{100} \times \$120)[/tex]

                                                          = $120 - $18 = $102

So, the new cost of a printer = $111 + $96 + $102 = $309

Now, the percentage increase in the production cost of the printer is given by;

      % increase =  [tex]\frac{\text{Net increase in the cost of printer}}{\text{Original cost of printer}} \times 100[/tex]

                         =  [tex]\frac{\$309- \$300}{\$300} \times 100[/tex]

                         =  3%

Hence, the percentage increase in the production cost of the printer is 3%.

Answer:

The probability is 0.31084

Step-by-step explanation:

We can calculate this probability using the z-score route.

Mathematically;

z = (x-mean)/SD/√n

Where the mean = 16, SD = 3 and n = 36

For 15.8, we have;

z = (15.8-16)/3/√36 = -0.2/3/6 = -0.2/0.5 = -0.4

For 16.2, we have

z = (16.2-16)/3/√36 = 0.2/3/6 = 0.2/0.5 = 0.4

So the probability we want to calculate is;

P(-0.4<z<0.4)

We can get this using the standard normal distribution table;

So we have;

P(-0.4 <z<0.4) = P(z<-0.4) - P(z<0.4)

= 0.31084

Answer:

We conclude that no more than 10% of its microwaves need repair during the first five years of use.

Step-by-step explanation:

We are given that a maker of microwave ovens advertises that no more than 10% of its microwaves need repair during the first 5 years of use.

In a random sample of 50 microwaves that are 5 years old, 12% needed repairs.

Let p = population proportion of microwaves who need repair during the first five years of use.

So, Null Hypothesis, [tex]H_0[/tex] : p [tex]\leq[/tex] 10%      {means that no more than 10% of its microwaves need repair during the first five years of use}

Alternate Hypothesis, [tex]H_A[/tex] : p > 10%     {means that more than 10% of its microwaves need repair during the first five years of use}

The test statistics that will be used here is One-sample z-test for proportions;

                        T.S.  =  [tex]\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]  ~ N(0,1)

where, [tex]\hat p[/tex] = sample proportion of microwaves who need repair during the first 5 years of use = 12%

           n = sample of microwaves = 50

So, the test statistics =  [tex]\frac{0.12-0.10}{\sqrt{\frac{0.10(1-0.10)}{50} } }[/tex]

                                    =  0.471

The value of z-test statistics is 0.471.

Now, at a 0.04 level of significance, the z table gives a critical value of 1.751 for the right-tailed test.

Since the value of our test statistics is less than the critical value of z as 0.471 < 1.751, so we have insufficient evidence to reject our null hypothesis as the test statistics will not fall in the rejection region.

Therefore, we conclude that no more than 10% of its microwaves need repair during the first five years of use.

Answer: C. 41

Step-by-step explanation:

[tex]-\left(31+2\right)+7^2-\left(-5^2\right)[/tex]

[tex]=-33+7^2-\left(-5^2\right)[/tex]

[tex]\left(-5^2\right)=-25[/tex]

[tex]=-33+7^2-\left(-25\right)[/tex]

[tex]7^2=49[/tex]

[tex]=-33+49-\left(-25\right)[/tex]

[tex]-33+49=16[/tex]

[tex]=16-\left(-25\right)[/tex]

[tex]\mathrm{Apply\:rule\:}-\left(-a\right)\:=\:+a[/tex]

[tex]16+25=41[/tex]

Answer:

150000 doctoral degrees

Step-by-step explanation:

From the graph attached,

At t = 0 represents the year 2005 and each unit of y-axis represents 10000 degrees.

Number of approximate number of degrees awarded to women in 2008,

f'(3) = 2.25

Similarly, number of doctoral degrees awarded from 2009 to 2013 are,

f'(4) = 2.5

f'(5) = 2.5

f'(6) = 2.5

f'(7) = 2.5

f'(8) = 2.75

Total number of degrees awarded to women from the start of 2008 to the start of 2014 = (2.25 + 2.5 + 2.5 + 2.5 + 2.5 + 2.75 ) × 10000

= 15 × 10000

= 150000

Answer:

Option A (repeated measures design) is the correct option.

Step-by-step explanation:

The other three options are not related to the given instance. So that alternative A would be the correct choice.

Answer:

Step-by-step explanation:

[tex](1+6i)-(7+3i) ?\\Group\:the\:real\:part\:and\:the\:imaginary\\\:part\:of\:the\:complex\:number\\\left(a+bi\right)\pm \left(c+di\right)=\left(a\:\pm \:c\right)+\left(b\:\pm \:d\right)i\\=\left(1-7\right)+\left(6-3\right)i\\1-7=-6\\6-3=3\\=-6+3i[/tex]

Answer:

34 grams

Step-by-step explanation:

If the remaining sample has 30.26 grams of radioactive substance, and 11% of it decayed, that means that 30.26 grams is 89% of the original. Let the original be x.

30.26=0.89x

Multiply both by one hundred

3026=89x

Divide both by 89

34=x

x=original, so the original was 34 grams.

Answer:

Step-by-step explanation:

a. What test is appropriate for this analysis?

A Chi-square  test of independence is appropriate for this analysis because it is used to compare two variables or testing relationship on categorical variables.  

b. State the null hypothesis:

The null hypothesis is the default hypothesis

[tex]\mathtt{H_o:}[/tex] There is no particular preference for any brand of toothpaste among students.

c. State the alternative hypothesis:

The alternative hypothesis is the research hypothesis which comes in place to challenge the validity of the null hypothesis.

[tex]\mathtt{H_a:}[/tex] There is particular preference for brands of toothpaste among students.

d. Find the critical value:

degree of freedom = n-1

degree of freedom = 3 - 1

degree of freedom = 2

At the level of significance ∝ = 0.05

The confidence interval = 0.95 and degree of freedom = 2, the critical value from the chi-square distribution table = 5.991

e. Calculate the test statistic:

Using the chi square test statistics; we have the following:

Crest             Colgate            Aquafresh            Total

24                        13                     8                      45

Since we have three brands. Then, for each brand, the expected value

= Total /3

= 45/3

=15

Thus:

Chi -square [tex]\mathtt{X^2 = \dfrac{(observed \ value - expected \ value)^2}{expected \ value}}[/tex]

[tex]\mathtt{X^2 = \dfrac{(24 - 15)^2}{15} + \dfrac{(13 - 15)^2}{15} + \dfrac{(8 - 15)^2}{15} }[/tex]

[tex]\mathtt{X^2 = \dfrac{81}{15} + \dfrac{4}{15} + \dfrac{49}{15} }[/tex]

[tex]\mathtt{X^2 = \dfrac{81+4+49}{15}}[/tex]

[tex]\mathtt{X^2 = \dfrac{134}{15}}[/tex]

[tex]\mathtt{X^2 =8.93 }[/tex]

f. Make a decision:

Since the chi-square value is greater than the critical value , we reject the null hypothesis and conclude that the students have  particular preference for brands of toothpaste.

Answer:

155 + 165.3 - 65.5 - 23.25 - 26.45

Step-by-step explanation:

She had $155 dollars in the starting = +155

She withdrew $65.5 = -65.5

She withdrew another $23.25 = -23.25

She withdrew another $26.45 = -26.45

The deposited $165.3 = +165.3

The expression looks like:

155 + 165.3 - 65.5 - 23.25 - 26.45

We could simplify the expression:

155 + 165.3 - 65.5 - 23.25 - 26.45

=> 320.3 - 88.75 - 26.45

=> 320.3 -115.2

=> 205.1

At the end of the week, she had a total of $205.10.

Midpoint formula: (x1 + x2)/2 , (y1 + y2)/2

Midpoint AB = (3 +-5)/2, (3 + 7)/2 = -2/2 , 10/2 = (-1,5)

Midpoint CD = (2 +9)/2, (11 + -2)/2 = (11/2,9/2)

Answer:

A; The first choice.

Step-by-step explanation:

We have the equation [tex]x^4+6x^2+5=0[/tex] and we want to solve using u-substitution.

When solving by u-substitution, we essentially want to turn our equation into quadratic form.

So, let [tex]u=x^2[/tex]. We can rewrite our equation as:

[tex](x^2)^2+6(x^2)+5=0[/tex]

Substitute:

[tex]u^2+6u+5=0[/tex]

Solve. We can factor:

[tex](u+5)(u+1)=0[/tex]

Zero Product Property:

[tex]u+5=0\text{ and } u+1=0[/tex]

Solve for each case:

[tex]u=-5\text{ and } u=-1[/tex]

Substitute back u:

[tex]x^2=-5\text{ and } x^2=-1[/tex]

Take the square root of both sides for each case. Since we are taking an even root, we need plus-minus. Thus:

[tex]x=\pm\sqrt{-5}\text{ and } x=\pm\sqrt{-1}[/tex]

Simplify:

[tex]x=\pm i\sqrt{5}\text{ and } x=\pm i[/tex]

Our answer is A.

Answer:

84 pounds

Step-by-step explanation:

If 1/3 of a book is equal to 28 pounds then 28*3 will give you your answer

Answer:

42 minutes

Step-by-step explanation:

if you are asking how long it takes Chelsea to play her tuba then you do:       67 - 25 = 42

Answer:

Step-by-step explanation:

jhngjnh

Answer:

(-2,-7)

Step-by-step explanation:

because it's reflected across the x-axis, only the y-intercept will change

Answer:

(-2,-7)

Step-by-step explanation:

All you have to do is draw a graph and draw the point across the x axis in the same row and same distance from the x axis.The distance is 7 so you just change it to -7.

Answer:

[tex]1(x {}^{8} + x {}^{4} + 1)[/tex]

Step-by-step explanation:

[tex]x {}^{8} + {x}^{4} + 1 =1( x {}^{8} + x {}^{2} + 1)[/tex]

Hope this helps ;) ❤❤❤

Let me know if there is an error in my answer.

Answer:

[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}) = \frac{11}{20}[/tex]

Step-by-step explanation:

Given

[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100})[/tex]

Required

Simplify

For clarity, group the expression in threes

[tex]((\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

Evaluate the first group [Divide 8 by 4]

[tex]((\frac{3}{1})*(\frac{2}{9})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 9 by 3]

[tex]((\frac{1}{1})*(\frac{2}{3})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex]((\frac{2}{3})*(\frac{15}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 15 by 3]

[tex]((\frac{2}{1})*(\frac{5}{16}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 16 by 2]

[tex]((\frac{1}{1})*(\frac{5}{8}))*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex](\frac{5}{8})*((\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

Evaluate the second group [Divide 35 and 25 by 5]

[tex](\frac{5}{8})*((\frac{24}{5})*(\frac{7}{36})*(\frac{48}{49}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 49 by 7]

[tex](\frac{5}{8})*((\frac{24}{5})*(\frac{1}{3})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[Divide 24 by 3]

[tex](\frac{5}{8})*((\frac{8}{5})*(\frac{1}{1})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex](\frac{5}{8})*((\frac{8}{5})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

Merge the first and second group

[tex]((\frac{5}{8})*(\frac{8}{5})*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex](1*(\frac{4}{7}))*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

[tex](\frac{4}{7})*((\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}))[/tex]

Evaluate the last group [Divide 99 by 9]

[tex](\frac{4}{7})*((\frac{63}{64})*(\frac{80}{9})*(\frac{11}{100}))[/tex]

[Divide 63 by 9]

[tex](\frac{4}{7})*((\frac{7}{64})*(\frac{80}{1})*(\frac{11}{100}))[/tex]

[Divide 64 and 80 by 8]

[tex](\frac{4}{7})*((\frac{7}{8})*(\frac{10}{1})*(\frac{11}{100}))[/tex]

[Divide 10 and 4 by 2]

[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{5}{1})*(\frac{11}{100}))[/tex]

[Divide 100 by 5]

[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{1}{1})*(\frac{11}{20}))[/tex]

[tex](\frac{4}{7})*((\frac{7}{4})*(\frac{11}{20}))[/tex]

[tex](\frac{4}{7})*(\frac{7}{4})*(\frac{11}{20})[/tex]

[tex]1*(\frac{11}{20})[/tex]

[tex]\frac{11}{20}[/tex]

Hence;

[tex](\frac{3}{4})*(\frac{8}{9})*(\frac{15}{16})*(\frac{24}{25})*(\frac{35}{36})*(\frac{48}{49})*(\frac{63}{64})*(\frac{80}{81})*(\frac{99}{100}) = \frac{11}{20}[/tex]

Answer:

The acceleration on the unit is 0.667 m/s^2

The tension on the draw-bar is 13.34 kN

Step-by-step explanation:

The mass of the truck = 10 Mg = 10 x 10^3 kg

The mass of the trailer = 20 Mg = 20 x 10^3 kg

Tractive force from the truck = 20 kN = 20 x 10^3 N

The total mass of the unit = 10 Mg + 20 Mg = 30 Mg = 30 x 10^3 kg

The tractive force on the unit will produce an acceleration that is given as

F = ma

where

F is the tractive = 20 x 10^3 N

m is the mass of the unit = 30 x 10^3 kg

a is the acceleration of the unit = ?

substituting into the equation

20 x 10^3 = 30 x 10^3 x a

a = (20 x 10^3)/(30 x 10^3) = 0.667 m/s^2

the tension on the draw-bar T is gotten from considering only the mass that is pulled by the draw-bar which is 20 Mg

The acceleration on the unit = 0.667 m/s^2

The drawn mass = 20 Mg = 20 x 10^3 kg

The tension on the draw bar = ma = 20 x 10^3 x 0.667 = 13340 N

= 13.34 kN

The acceleration is 0.00067m/s^2, while the tension on the horizontal bar is 13.4 N

The given parameters are:

[tex]\mathbf{m = 10Mg}[/tex] -- mass of the truck

[tex]\mathbf{M = 20Mg}[/tex] -- mass of the trailer

[tex]\mathbf{F_T = 20kN}[/tex] --- tractive force

Start by calculating the total mass

[tex]\mathbf{M_T = m + M}[/tex]

So, we have:

[tex]\mathbf{M_T = 10Mg + 20Mg}[/tex]

[tex]\mathbf{M_T = 30Mg}[/tex]

Convert to kilograms

[tex]\mathbf{M_T = 30 \times 10^3kg}[/tex]

[tex]\mathbf{M_T = 30000 kg}[/tex]

Force is calculated as:

[tex]\mathbf{F =ma}[/tex]

So, we have:

[tex]\mathbf{20kN =30000kg \times a}[/tex]

Divide both sides by 30000

[tex]\mathbf{a = 0.00067ms^{-2}}[/tex]

The tension on the horizontal bar (i.e. the 20 Mg trailer) is:

[tex]\mathbf{T=ma}[/tex]

So, we have:

[tex]\mathbf{T=20Mg \times 0.00067ms^{-2}}[/tex]

Rewrite as:

[tex]\mathbf{T=20 \times 10^3 kg \times 0.00067m/s}[/tex]

[tex]\mathbf{T=13.4N}[/tex]

Hence, the acceleration is 0.00067m/s^2, while the tension on the horizontal bar is 13.4 N

Read more about force and acceleration at:

https://brainly.com/question/20511022

Answer:

6

Step-by-step explanation:

Hello, please consider the following.

[tex](4+x)^2=4^2+2\cdot 4\cdot x+x^2=16+\boxed{8}x+x^2\\\\\text{ ... and not ...}\\\\16+\boxed{4}x+x^2[/tex]

So the correct equation becomes.

[tex]x^2+64=16+8x+x^2\\\\8x=64-16=48\\\\\text{ we divide by 8 both sides of the equation.}\\\\x=\dfrac{45}{8}=6[/tex]

Hope this helps.

Do not hesitate if you need further explanation.

Thank you

Answer:

Error : The expression ( 4 + x )² was expanded incorrectly.

Correct Solution : x = 6

Step-by-step explanation:

The planning of the solution is correct, by Pythagorean Theorem you can say that PQ² + QO² = PO², and hence through substitution x² + 8² = ( 4 + x )². Let's look into the calculations.

PQ² + QO² = PO²,

x² + 8² = ( 4 + x )²,

x² + 8² = 16 + 8x + x²,

64 = 16 + 8x,

48 = 8x,

x = 48 / 8 = 6, x = 6

As you can see, the only error in the calculations was expanding the expression ( 4 + x )². ( 4 + x )² = 4² + 2 [tex]*[/tex] 4 [tex]*[/tex] x + x² = 4² + 8x + x² = 16 + 8x + x², not 16 + 4x + x².

Answer:

Step-by-step explanation:

Formula for area of a triangle:

Height x Base /2

Base (b) = h +4

Height = h

h + 4 x h /2 = 30cm

=> h +4 x h = 60

=> h+4h =60

=> 5h = 60

=> h = 12

Height = 12

Base = 12 +4 = 16

Answer:

add 8x to both sides

Step-by-step explanation:

5-8x<2x+3

first step, subtract 3 from both sides:

2-8x<2x

second step,?

2<?x

so you need to add 8x first

Answer:

The work done by the force is 42.4 Joules

Step-by-step explanation:

The force F = 7 pounds

angle to the horizontal that the force acts ∅ = 30°

The object is moved a distance d = 7 feet

The coordinate  (0 comma 0 )to (7 comma 0 ), indicates that the movement started from the origin, and is along the x-axis.

The work done by this force = F cos ∅ x d

==> 7 cos 30° x 7

==> 7 x 0.866 x 7 = 42.4 Joules

Answer:

Shift the graph of y = x2 right 6 units.

Answer:

Step-by-step explanation:

The summary of the given data includes;

sample size for the first school [tex]n_1[/tex] = 42

sample size for the second school [tex]n_2[/tex]  = 34

so 16 out of 42 i.e [tex]x_1[/tex] = 16 and 18 out of 34 i.e [tex]x_2[/tex] = 18 have ear infection.

the proportion of students with ear infection Is as follows:

[tex]\hat p_1 = \dfrac{16}{42}[/tex] = 0.38095

[tex]\hat p_2 = \dfrac{18}{34}[/tex]  =  0.5294

Since this is a two tailed test , the null and the alternative hypothesis can be computed as :

[tex]H_0 :p_1 -p_2 = 0 \\ \\ H_1 : p_1 - p_2 \neq 0[/tex]

level of significance ∝ = 0.05,

Using the table of standard normal distribution, the value of z that corresponds to the two-tailed probability 0.05 is 1.96. Thus, we will reject the null hypothesis if the value of the test statistics is less than -1.96 or more than 1.96.

The test statistics for the difference in proportion can be achieved by using a pooled sample proportion.

[tex]\bar p = \dfrac{x_1 +x_2}{n_1 +n_2}[/tex]

[tex]\bar p = \dfrac{16 +18}{42 +34}[/tex]

[tex]\bar p = \dfrac{34}{76}[/tex]

[tex]\bar p = 0.447368[/tex]

[tex]\bar p + \bar q = 1 \\ \\ \bar q = 1 -\bar p \\ \\\bar q = 1 - 0.447368 \\ \\\bar q = 0.552632[/tex]

The pooled standard error can be computed by using the formula:

[tex]S.E = \sqrt{ \dfrac{ \bar p \bar q}{ n_1} + \dfrac{\bar p \bar p}{n_2} }[/tex]

[tex]S.E = \sqrt{ \dfrac{ 0.447368 * 0.552632}{ 42} + \dfrac{ 0.447368 * 0.447368}{34} }[/tex]

[tex]S.E = \sqrt{ \dfrac{ 0.2472298726}{ 42} + \dfrac{ 0.2001381274}{34} }[/tex]

[tex]S.E = \sqrt{ 0.01177284105}[/tex]

[tex]S.E = 0.1085[/tex]

The test statistics is ;

[tex]z = \dfrac{\hat p_1 - \hat p_2}{S.E}[/tex]

[tex]z = \dfrac{0.38095- 0.5294}{0.1085}[/tex]

[tex]z = \dfrac{-0.14845}{0.1085}[/tex]

z = - 1.368

Decision Rule: Since the test statistics is greater than the rejection region - 1.96 , we fail to reject the null hypothesis.

Conclusion: There is insufficient evidence to support the claim that a difference exists between the proportions of students who have ear infections at the two schools