-(9m + 7) algebra question please help me!!

Answer: V=πr2h

3

Step-by-step explanation: Your spelling is a bit off

Critical value with 35 degrees of freedom at 0.05 Los = 2.030

Given information,

Departmental liabilities have a mean and standard deviation that are unknown for the population. What t-score should be used to establish a 95% confidence interval estimate for the population mean if a random sample of 36 departments is utilized to estimate the mean liabilities?

Here,

By assuming a 95% confidence interval and by considering the t-value we have,

Therefore;

Critical value at 0.05 Los with 35 degrees of freedom =2.030

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If Shen produced and Manuel consumed one additional bottle of water, total surplus would be $2

The law of supply and demand combines two fundamental economic ideas that describe how changes in the price of a resource, commodity, or product affect its supply and demand.

Supply grows as the price rises, but demand drops. In contrast, as the price falls, supply is constrained and demand is increased.

Demand and supply levels for different prices can be displayed as curves on a graph. The intersection of these curves symbolizes the process of price discovery in the marketplace, as it indicates the balance, or market-clearing price, at which demand equals supply.

According to the law of demand, demand for a good or resource will decrease as its price increases and increase as its price decreases.

How to solve?

Price ------ Quantity Supplied ----- Quantity Demanded

2 --------------------- 1 ------------------------- 3

4 --------------------- 2 ------------------------ 2

6 --------------------- 3 ------------------------ 1

4 (quantity supplied = quantity demanded = 2)

4 ($7>$4 and $5>$4, $3<$4<$5, area above price line but under curve if price line was at $4, $7-$4=$3, $5-$4=$1, $3+$1=$4); 4 ($1<$4 and $3<$4, $3<$4<$5, area below price line and under curve if price line at $4, $4-$1=$3, $4-$3=$1, $3+$1=$4); 8 ($4 from consumer surplus + $4 from producer surplus)

fall (equilibrium = 2, 2-1=1, Shen: quantity supplied = 1 -> $2, Manuel: quantity demanded = 1 -> $6, total surplus = producer surplus + consumer surplus = ($4-$1)+($7-$4)=$6, $8>$6)

rise (equilibrium = 2, 2+1=3, Manuel: quantity demanded = 3 -> $2, Shen: quantity supplied = 3 -> $6, total surplus = consumer surplus + producer surplus = ($3-$2)+($7-$6)=$2, $8>$2, $6>$2)

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The values of a, b and c in the mixed fractions are 1, 5 and 9 respectively.

The fractions are mixed fractions . let's convert it to improper fraction be we simplify its.

A mixed fraction is a whole number and a proper fraction combined. Improper fractions are defined as fractions for which the numerator is greater than the denominator.

Therefore, let's solve the fractions as follows:

[tex]3\frac{1}{2}[/tex] ÷ [tex]2\frac{1}{4}[/tex]

[tex]3\frac{1}{2} = \frac{7}{2}[/tex]

[tex]2\frac{1}{4} = \frac{9}{4}[/tex]

Therefore,

[tex]\frac{7}{2} (\frac{4}{9} ) = \frac{28}{18} = 1\frac{10}{18} = 1\frac{5}{9}[/tex]

Hence,

a = 1

b = 5

c = 9

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The paper's smallest dimensions are height=8.660in and width=17.321in.

In physics and mathematics, the dimension of space is the smallest set of coordinates needed to specify any point within a mathematical space.

A line has a dimension of one because only one coordinate is needed to specify a point on a line, such as a point at 5 on a number.

So, we have:

Area = 150in²

Side margin = 1 in

Top and Bottom Margin = 2 in

To obtain the smallest dimensions, calculate as follows:

The printed material's area is:

Area = l*w

A1 = h*w

A1 should be changed to 150:

150 = h*w

h = 150/w

The complete paper's topic is:

Area = l*w

A2 = (w+2+2)*(H+1+1)

A2 = (w+4)*(H+2)

For H, substitute 150/W:

A2 = (w+4)*(150/w+2)

A2 = w(150/w+2) + 4(150/w+2)

A2 = 150 + 2w + 600w + 8

A2 = 2w = 600/w + 8 + 150

A2 = 2w = 600/w + 158

Calculate a difference with regard to w and set the outcome to 0:

A'2 = 2-600/w²+0

A'2 = 2-600/w²

0 = 2 - 600/w²

2 = 600/w²

2*w² = 600

w² = 600/2

w² = 300

w = 17.321

Remember that:

h = 150/w

h = 150/17.321

h = 8.660

Therefore, the paper's smallest dimensions are height=8.660in and width=17.321in.

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Tn the given situation, the appro[riate null hypotheses and alternative hypotheses are: H₀: μ₁ ≥ μ₂a and H₁: μ₁ < μ₂

An assertion used in statistical inference experiments is known as the alternative hypothesis.

It is indicated by Ha or H1 and runs counter to the null hypothesis.

Another way to put it is that it is only a different option from the null.

An alternative theory in hypothesis testing is a claim that the researcher is testing.

The initial assertion that forms the null hypothesis frequently rests on research from the past or expert knowledge.

According to the alternative hypothesis, a population parameter is either less, larger, or different from the value predicted by the null hypothesis.

So, the new production method supposedly lowers the mean operating cost per hour.

The following are the hypothesis when comparing the average production costs of the new production method (μ₁) with the old production method (μ₂):

H₀: μ₁ ≥ μ₂

H₁: μ₁ < μ₂

Therefore, in the given situation, the appro[riate null hypotheses and alternative hypotheses are: H₀: μ₁ ≥ μ₂a and H₁: μ₁ < μ₂

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After approximately 113 minutes, there will be only 40 bacteria remaining.

Half-life is defined as the time it will took a substance to reduce to half of its initial amount. The half-life of a substance is determined with respect to the rate of the reaction. It can be calculated using the formula below.

N(t) = N₀(1/2)^(t/T)

where N(t) = amount of substance remaining after time t

N₀ = initial amount of substance

t = time

T = half-life

If the sample begins with 500 bacteria, and after 16 minutes there are 350 bacteria, then

N₀ = 500, and N(t) = 350 when t = 16 minutes

Substitute the given values and solve for the half-life.

N(t) = N₀(1/2)^(t/T)

350 = 500(1/2)^(16/T)

T = 31.09 minutes

Using the formula, solve for t when n(t) = 40.

N(t) = N₀(1/2)^(t/T)

40 = 500(1/2)^(t/31.09)

t = 113.30 minutes

t ≅ 113 minutes

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The total cost for a team of 10 player in a baseball league is $978.5.

For the stated question,

The cost of each equipment for each player is-

Thus,

Total cost for each player =  $50. 95  + $20. 95  +  $25. 95

Total cost for each player = $97.85

Total number of player in a baseball league = 12

Total cost for team = 10 x $97.85

Total cost for team = $978.5

Thus, the total cost for a team of 10 player in a baseball league is $978.5.

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Answer:

-3

Step-by-step explanation:

6x + 5 = 0 - there are no solutions...

Answer:

Step-by-step explanation:

-1/3

Answer:

Positive seven.

Step-by-step explanation:

Go past zero to normal seven.

Answer:positive 7

Step-by-step explanation:

The solution for y in the expression is y = -53 and y = 63

From the question, the expression is given as

y = 5 ± 58

We start by expanding the expression

So, we have the following representation

y = 5 - 58 and y = 5 + 58

Evaluate the sum and the difference of the like terms in the above expression

So, we have the following representation

y = -53 and y = 63

Hence, the solution is

y = -53 and y = 63

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Answer:15 nickels 22 dimes

Step-by-step explanation:

Answer :not fully correct

cStep-by-step explanation:

Answer:

A

Step-by-step explanation:

describes a good place

Answer:7:20

Step-by-step explanation:

On solving the question, we got that -  [tex]P(t) = P_{100e^{-0.058t} }[/tex] grams of radioactive compound will remain after 2 hours.

Chemicals known as radionuclides (also known as radioactive materials) have unstable atoms in their center. Changes in the nucleus help them maintain stability, spontaneous fission, emission of alpha particles, or conversion of neutrons to protons or the reverse.

Future value is the issue at hand. If weight is measured in grams and time is measured in years, then [tex]P(t) = P_{0e^{-rt} }[/tex] with [tex]P_{0} = 100[/tex] and r = 0.058

Therefore,

[tex]P(t) = P_{100e^{-0.058t} }[/tex]

Will remain after 2 hours.

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Answer:

it would take him 4/7 of an hour

Step-by-step explanation:

Answer:

80

usually estimate is rounded 1 significant figure, 0.9 is rounded to 1, and 82 is rounded to 80.

1*80 is 80

(1) It will take 17.8 hours for Sam's family to get to Chicago, Illinois .

(2) Rohan's family will take 17.9 hours to arrive in Chicago, Illinois .

In the question ,

it is given that ,

For Sam's Family ,

total distance to be covered is = 1157  miles ,

the average speed is given to be = 65 mph .

By the formula of Speed , Distance and Time ,

we get ,

65 = 1157/t

t = 1157/65

t = 17.8 hours .

For Rohan's family ,

total distance to be covered is = 1250  miles ,

the average speed is given to be = 70 mph ,

By the formula of Speed , Distance and Time ,

we get ,

70 = 1250/t

t = 1250/70

t = 17.9 hours

Therefore , Sam's family will take 17.8 hours and Rohan's family will take 17.9 hours .

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Answer:

Step-by-step explanation:

The ratio formula is given as a:b = a/b for any two quantities, let's say a and b. Given that a and b are separate amounts for two quantities, the combined total quantity is denoted by (a + b).

There are 48 lollipops and 6 candy bars for a gift bag.

Then the possible ratios of lollipops to candy bars are:

( 48 / 6 ) ÷ ( 2 / 2) = ( 24 / 3 )

( 48 / 6 ) ÷ ( 3 / 3) = ( 16 / 2 )

( 48 / 6 ) ÷ ( 6 / 6) = ( 8 / 1 )

There are 48 lollipops for every 6 candy bars.

There are 24 lollipops for every 3 candy bars.

There are 16 lollipops for every 2 candy bars.

There are 8 lollipops for every 1 candy bar.

The statement is true that is the standard error of estimate measures the accuracy of a prediction.

Given that,

We have a statement that is the standard error of estimate measures the accuracy of a prediction.

We have to find is the statement is true are false.

We know that,

The SE formula is used to determine how accurately a sample represents a population. The sample mean's departure from the specified population is provided as follows,

Formula for standard error

S.E=S/√n

Where, n is the number of observations and S is the standard deviation.

Therefore, The statement is true that is the standard error of estimate measures the accuracy of a prediction.

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Using the definition of covariance, Cov(Y(1)Y(2))= 0. From the given information the two variables Y(1), and Y(2), are dependent and, p(y(1)y(2)) ≠ P(y(1))p(y(2)). So, it can be concluded that Uncorrelated variables need not be independent.

Given joint probability function of Y(1) and Y(2), is

p(Y(1)*Y(2)) = 1/3, for (y(1)*y(2)) = (- 1, 0), (0, 1), (1, 0)

So Y(1), takes random variable -1,0,1.

And Y(2) takes random variable 0,1.

Y(2)|Y(1)      -1               0             1

0                  1/3            0             1/3

1                   0               1/3           0

From the table:

The marginal probability function of Y(1) is;

P(Y(1)=y(1)) = ∑P(Y(1)=y(1)), y(1)= -1,0,1. That is;

Y(1)=y(1)                      -1                   0               1

P(Y(1)=y(1))                  1/3                 1/3             1/3

From the definition of expectation,

E[y(1)]= [tex]\Sigma_{y_{1}}[/tex] y(1)P(Y(1)=y(1))

E[y(1)]= -1(1/3)+0(1/3)+1(1/3)

E[y(1)]= -1/3+0+1/3

E[y(1)]= 0

The marginal probability mass function of y(2) is

P(Y(2)=y(2)) = ∑P(y(1)y(2)), y(2)= 0,1. That is;

Y(2)=y(2)                    0               1

[tex]P_{Y(2)}[/tex](y(2))                  2/3             1/3

Now E[y(2)]= [tex]\Sigma_{y_{2}}[/tex] y(2)P(Y(2)=y(2))

E[y(2)]= 0(2/3)+1(1/3)

E[y(2)]= 1/3

And

E[y(1)y(2)]= [tex]\Sigma_{y_{1}}\Sigma_{y_{2}}[/tex] y(1)y(2)P(y(1)y(2))

E[y(1)y(2)]= -1(0)(1/3)+0(1)(1/3)+1(0)(1/3)

E[y(1)y(2)]= 0

From the definition of covariance.

The Cov(Y(1)Y(2))= E[Y(1)Y(2)]-E[Y(1)E[Y(2)]

Cov(Y(1)Y(2))= 0-1/3 (0)

Cov(Y(1)Y(2))= 0

The Cov(Y(1)Y(2))=0, implies that the two variables are uncorrelated. But from the given information the two variables Y(1), and Y(2), are dependent and, p(y(1)y(2)) ≠ P(y(1))p(y(2))

Hence, it can be concluded that Uncorrelated variables need not be independent.

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The value of t while considering a right-tailed test (upper tail) and a sample size of 40 at the 95% confidence level is ±1.685.

Given:

Consider a right-tailed test (upper tail) and a sample size of 40 at the 95% confidence level.

we are asked to determine the t value:

The directional hypothesis is:

Hₐ : µ > µ₀

The sample size is, n = 40.

The significance level is:

α = 1-0.95 = 0.05

Use the t-table to compute the right-tailed t-critical value as follows:

tₐ,₍ₙ₋₁₎ = t₀.₀₅,₍₄₀₋₁₎

        = t₀.₀₅,₃₉

        ≈ t₀.₀₅,₄₀

        = 1.684

        ≈ 1.685

The value is positive since the directional hypothesis suggests that the test is right tailed.

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The solution of the algebraic equation for the cost of downloading a song for members is x = $0.95

Given that:

1. Peggy pays an annual fee of $10.75. Last year she downloaded 105 songs

2. Non-members pay $0.35 more per song download than members

3. Daniel is not a member of the club and downloaded 85 songs last year

Let x represent the cost of downloading a song for members

Thus, the cost of a song for Peggy will be x

From statement 2: the cost of a song for Daniel will be x+0.35

The algebraic equation for the total amount for Peggy will be:

A = 105x + 10.75

The algebraic equation for the total amount for Daniel will be:

A = 85(x+0.35)

If Daniel and Peggy both paid the same total amount, we can say:

105x + 10.75 = 85(x+0.35)

105x + 10.75 = 85x + 29.75

105x - 85x = 29.75 - 10.75

20x = 19

x = 19/20

x = 0.95

Thus, the cost of a song for member is $0.95

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Answer: 72

Step-by-step explanation: because 8 x 12 = 96  and 3 x 8 = 24  

96 - 24 = 72

The rate at which lemons being used in lemons per cup of lemonade is 0.25 lemons per cup

A ratio compares values. A ratio says how much of one thing there is compared to another thing.

Rate of Change

Rate of change is the value at which the rest of the time.

Let the rate of lemon per cup of lemonade be x

Total lemon = 4

For a cup of lemonade =16

Rate of lemon per cup of lemonade

x = 4/16

x= 0.25.

Therefore, the required rate of lemon per cup of lemonade is 0.25 lemons per cup

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The values of A, B, C and D are 55.04ft² ,9184.50ft , 75.36ft and 38.46ft², the A(C-D)+B= 11214.00

Area of a circle is the region occupied by the circle in a two-dimensional plane. It can be determined easily using a formula, A = πr2.

Circumference of a circle is the perimeter of a circle. It can be calculated as C= πd or 2πr

calculating A, the area of the circle is 3.14×8²= 200.96

the area of the square= l×l= 16×16= 256

area of the shaded part = 256-200.96= 55.04ft²

calculating B, since the maintenance worker moved round 3 times

3C = 3πd= 3.14×975×3 = 9184.50ft

calculating C

circumference of 3 bowls

= 3πd= 3×3.14×8= 75.36ft

calculating D

area= πr²= 3.14×3.5²

= 38.46ft²

therefore A(C-D)+B= 55(75.36-38.46)+9184.50

= 2029.5+9184.5

= 11214.00

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Step-by-step explanation:

"with replacement". that means any pulled ball is returned into the bag, and we have the full 7+3 = 10 balls to pull from every time.

remember, a probability is always

desired cases / totally possible cases

(a) 2 back will be red. I understand this, that exactly 2 balls will be red (not fewer, not more).

to get exactly 2 red balls we get 2 red balls and 4 black balls.

the probability for such a single event is

7/10 × 7/10 × 3/10 × 3/10 × 3/10 × 3/10 = 49×81/1000000 =

= 0.003969

because the probably to get a red ball (one from 7 out of 10) is 7/10.

the probability to get a black ball is 3/10.

how many different such single events are possible ?

as many as there are combinations to pick 2 out of 6 :

C(6, 2) = 6! / (2! × (6-2)!) = 6! / (2 × 4!) =

= 6×5 / 2 = 3×5 = 15

so, the probability to get exactly 2 red balls is

15 × 0.003969 = 0.059535 ≈ 0.05954

(b)

the probability of all 6 balls being black is

3/10 × 3/10 × 3/10 × 3/10 × 3/10 × 3/10 =

= 3⁶/1000000 = 0.000729 ≈ 0.00073

(c)

"at least 4 black balls" means exactly 4, or exactly 5, or exactly 6 balls are black.

that means we need to calculate all 3 probabilities and add them (non-overlapping "or"-cases means addition of probabilities, "and" cases mean multiplication as above).

to get exactly 4 black balls (= 4 black balls and 2 red balls) is the same as (a) : 0.059535

to get exactly 5 black balls (= 5 black balls and 1 red ball) in one specific constellation is

3⁵/10⁵ × 7/10 = 243×7/1000000 = 0.001701

and we have C(6, 5) possible constellations

C(6, 5) = 6! / (5! × (6-5)! = 6! / 5! = 6

the total probability to get exactly 5 black balls is

6 × 0.001701 = 0.010206

to get exactly 6 black balls is (b) : 0.000729

the probability to get at least 4 black balls is

0.059535 + 0.010206 + 0.000729 =

= 0.07047