Lena is comparing offers from two banks on checking accounts that include debit cards. Bank A charges $20 monthly fee for a checking account and debit card, with unlimited transactions. Bank B charged a $5 monthly fee for a checking account and debit card, plus
$ 0.50 for each transaction.
Suppose Lena makes 35 transactions in a given month.

How much would she pay at each bank for the given month?
Bank A
Bank B

For the given month, which bank is cheaper and by how much?
Bank A. is cheaper than Bank B by $
or
Bank B is cheaper than Bank A by $​

Step-by-step explanation:

(f+g)(x) = f(x) + g(x)

= x/2-3 + 4x²+x+4

= ..........

Answer:

x=9/2

Step-by-step explanation:

Let's solve your equation step-by-step.

5x+4(−x−2)=−5x+2(x−1)+12

Step 1: Simplify both sides of the equation.

5x+4(−x−2)=−5x+2(x−1)+12

5x+(4)(−x)+(4)(−2)=−5x+(2)(x)+(2)(−1)+12 (Distribute)

5x+−4x+−8=−5x+2x+−2+12

(5x+−4x)+(−8)=(−5x+2x)+(−2+12) (Combine Like Terms)

x+−8=−3x+10

x−8=−3x+10

Step 2: Add 3x to both sides.

x−8+3x=−3x+10+3x

4x−8=10

Step 3: Add 8 to both sides.

4x−8+8=10+8

4x=18

Step 4: Divide both sides by 4.

4x/4=18/4

x=9/2

Answer:

Step-by-step explanation:

[tex]-3\left(x+1\right)=-3\left(x+1\right)-5\\\\\mathrm{Add\:}3\left(x+1\right)\mathrm{\:to\:both\:sides}\\\\-3\left(x+1\right)+3\left(x+1\right)=-3\left(x+1\right)-5+3\left(x+1\right)\\\\\mathrm{Simplify}\\\\0=-5\\\\\mathrm{The\:sides\:are\:not\:equal}\\\\\mathrm{No\:Solution}[/tex]

Answer: 22,103

Step-by-step explanation:

Compound interest is the interest calculated on the initial principal and the accumulated interest.

The amount in the account at the end of two years is $22,050.

Compound interest is the interest calculated on the initial principal and the accumulated interest.

We have,

Principal = $20,000

Rate = r = 5%

It is compounded yearly.

Time = t = 2 years.

The formula for the amount having compound interest:

A = P [tex]( 1 + \frac{r}{n} )^{nt}[/tex]

A = 20,000 [tex](1 + \frac{5}{100\times1})^{2\times1}[/tex]

A = 20,000 ( 1 + 5/100 )²

A = 20,000 ( 105/100 )²

A = (20,000 x 105 x 105) / (100 x 100)

A = 2 x 105 x 105

A = $22,050

Thus the amount in the account at the end of two years is $22,050.

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Answer:

1. cups of coffee sold

2.Y = 605.7 - 5.943x

3. -0.952

4. 70.84

Step-by-step explanation:

1. the dependent variable in this question is the cups of coffee sold

2. least square estimation line

Y = a+bx

we have y as the cups of coffee sold

x as temperature.

first we will have to solve for a and then b

∑X = 450

∑Y = 960

∑XY = 61600

∑X² = 35500

∑Y² = 221800

a = ∑y∑x²-∑x∑xy/n∑x²-(∑x)²

a = 960 * 35500-450*61600/6*35500-450²

a = 6360000/10500

= 605.7

b = n∑xy - ∑x∑y/n∑x²-(∑x)²

= 6*61600 - 450*960/6*35500 - 450²

= -5.943

the regression line

Y = a + bx

Y = 605.7 - 5.943x

3. we are to find correlation coefficient

r = n∑xy - ∑x∑y multiplied by√(n∑x²-(∑x)² * (n∑y² - (∑y)²)

= 6*61600 -960*450/√(6*35500 - 450²)*(6*221800 - 960²)

=-62400/√4296600000

= -62400/65548.5

= -0.952

4. we have to predict sales of a 90 degree day fro the regression line

Y = 605.7 - 5.943x

y = 605.7 - 5.943(90)

y = 605.7 - 534.87

= 70.84

Answer:

9  3  -7  -13

4  -4  11  8

0  9  2  -4

Step-by-step explanation:

9  3  -7  -13

4  -4  11  8

0  9  2  -4

Answer: 9  3  -7  -13

4  -4  11  8

0  9  2  -4

Step-by-step explanation:

[tex](f\circ g)(2)=\dfrac{1}{4\cdot2+9}=\dfrac{1}{17}\\\\(f+g)(2)=\dfrac{1}{2}+4\cdot2+9=\dfrac{1}{2}+17=\dfrac{1}{2}+\dfrac{34}{2}=\dfrac{35}{2}[/tex]

Answer:

box plot titled ACT Score with a minimum of 24, quartile 1 of 27, median of 28, quartile 3 of 30, and maximum of 35

Step-by-step explanation:

The scores of the students represented on the dot plot are:

1 dot => 24

3 dots => 26, 26, 26

3 dots => 27, 27, 27

5 dots => 28, 28, 28, 28, 28

3 dots => 30, 30, 30

3 dots => 32, 32, 32

1 dot => 35

Quickly, we can ascertain 3 values from these data points of which we can use to find out which box plot represents the dot plot data.

The minimum score = 24

The maximum score = 35

The median score is the 10th value, which is the middle value of the data point = 28

Therefore, we can conclude that: "box plot titled ACT Score with a minimum of 24, quartile 1 of 27, median of 28, quartile 3 of 30, and maximum of 35".

Answer:

The 99% confidence interval for estimating the population mean μ is ($60,112.60, $68087.40).

Step-by-step explanation:

The complete question is:

Salaries of 42 college graduates who took a statistics course in college have a​ mean, [tex]\bar x[/tex] of, $64, 100. Assuming a standard​ deviation, σ of ​$10​,016 construct a ​99% confidence interval for estimating the population mean μ.

Solution:

The (1 - α)% confidence interval for estimating the population mean μ is:

[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]

The critical value of z for 99% confidence interval is:

[tex]z_{\alpha/2}=z_{0.01/2}=z_{0.005}=2.57[/tex]

Compute the 99% confidence interval for estimating the population mean μ as follows:

[tex]CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex]

     [tex]=64100\pm 2.58\times\frac{10016}{\sqrt{42}}\\\\=64100+3987.3961\\\\=(60112.6039, 68087.3961)\\\\\approx (60112.60, 68087.40)[/tex]

Thus, the 99% confidence interval for estimating the population mean μ is ($60,112.60, $68087.40).

Answer:

less than

Step-by-step explanation:

If the normality requirement is not satisfied​ (that is, ​np(1​ - p) is not at least​ 10), then a​ 95% confidence interval about the population proportion will include the population proportion in​ _less than__ 95% of the intervals.

The confidence interval consist of all reasonable values of a population mean. These are value for which the null hypothesis will not be rejected.

So, let assume that If the 95%  confidence interval contains the value for the hypothesized mean, then the sample mean  is reasonably close to the hypothesized mean. The effect of this is that the p- value is going to be greater than 0.05, so we fail to reject the null hypothesis.

On the other hand,

If the 95%  confidence interval do not contains the value for the hypothesized mean, then the sample mean  is far away from the hypothesized mean. The effect of this is that the p- value is going to be lesser than 0.05, so we reject the null hypothesis.

Answer:

Step-by-step explanation:

Probability is the likelihood or chance that an event will occur.

Probability = expected outcome of event /total outcome

Since the population consists of 100 elements, the total outcome of event = 100.

If random sample of 20 element is drawn from the population, the expected outcome = 20

On the first selection, the probability of any particular element being selected = 20/100 = 1/5

Complete Question

The complete question is shown on the first uploaded image

Answer:

Option C is the correct option

Step-by-step explanation:

From the question we are told that

   The equation is  [tex]f (x, y , z ) = x^2 +y^2 + z^2[/tex]

    The constraint is  [tex]P(x, y , z) = x + y + z - 24 = 0[/tex]

Now using Lagrange multipliers  we have that  

   [tex]\lambda = \frac{ \delta f }{ \delta x } = 2 x[/tex]  

   [tex]\lambda = \frac{ \delta f }{ \delta y } = y[/tex]  

   [tex]\lambda = \frac{ \delta f }{ \delta z } = 2 z[/tex]

=>       [tex]x = \frac{ \lambda }{2}[/tex]

          [tex]y = \frac{ \lambda }{2}[/tex]

         [tex]z = \frac{ \lambda }{2}[/tex]

From the constraint  we have

      [tex]\frac{\lambda }{2} + \frac{\lambda }{2} + \frac{\lambda }{2} = 24[/tex]

=>   [tex]\frac{3 \lambda }{2} = 24[/tex]

=>   [tex]\lambda = 16[/tex]

substituting for x, y, z

=>   x =  8

=>  y =  8

=>   z =  8        

Hence

    [tex]f (8, 8 , 8 ) = 8^2 +8^2 + 8^2[/tex]

    [tex]f (8, 8 , 8 ) = 192[/tex]

Complete Question

A research center claims that ​30% of adults in a certain country would travel into space on a commercial flight if they could afford it. In a random sample of 700 adults in that​ country, ​34% say that they would travel into space on a commercial flight if they could afford it. At ​, is there enough evidence to reject the research center's claim

Answer:

Yes there is  sufficient evidence to reject the research center's claim.

Step-by-step explanation:

From the question we are told that

     The population proportion is  p = 0.30

      The sample proportion is  [tex]\r p = 0.34[/tex]

       The  sample size is  n = 700

The null hypothesis is  [tex]H_o : p = 0.30[/tex]

 The  alternative hypothesis is  [tex]H_a : p \ne 0.30[/tex]

Here we are going to be making use of  level of significance  =  0.05 to carry out this test

Now we will obtain the critical value of  [tex]Z_{\alpha }[/tex] from the normal distribution table , the value is  [tex]Z_{\alpha } = 1.645[/tex]

 Generally the test statistics is mathematically represented as

            [tex]t = \frac{ \r p - p }{ \sqrt{ \frac{ p (1-p)}{n} } }[/tex]

substituting values

              [tex]t = \frac{ 0.34 - 0.30 }{ \sqrt{ \frac{ 0.30 (1-0.30 )}{ 700} } }[/tex]

              [tex]t = 2.31[/tex]

Looking at the values of t  and  [tex]Z_{\alpha }[/tex] we see that [tex]t > Z_{\alpha }[/tex] hence the null hypothesis is rejected

 Thus we can conclude that there is  sufficient evidence to reject the research center's claim.

Answer:

(a). x = 80°

(b). x = 7.2 units

Step-by-step explanation:

Angle formed between the tangents from a point outside the circle measure the half of the difference of intercepted arcs.

(a). Here the intercepted arcs are,

    Measure of major arc = 360° - 100°

                                        = 260°

    Measure of minor arc = 100°

   x° = [tex]\frac{1}{2}[m(\text{Major arc})-m(\text{Minor arc})][/tex]

       = [tex]\frac{1}{2}(260-100)[/tex]

    x = 80°

(b). If a secant and tangent are drawn form a point outside the circle, then square of the measure of tangent is equal to the product of the measures of the secant segment and and its external segment.

x² = 4(4 + 9)

x² = 4 × 13

x² = 52

x = √52

x = 7.211 ≈ 7.2 units

Answer:

6.25

Step-by-step explanation:

2.5 *2.5=6.25

Answer:

6.25cm^2.

Step-by-step explanation:

To find the area of a square, you multiply the two sides, 2.5✖️2.5.

This gives the area of 6.25cm^2.

Hope this helped!

Have a nice day:)

There will be a total of 4 test scores including the final exam. To get a 70, the 4 tests need to equal 4 x 70 = 280 points , to be 79, they have to equal 4 x 79 = 316 points.

The 3 already done = 61 + 62 + 86 = 209 points.

The final exam needs to be between :

280 -209 = 71

316 -209 = 107. The answer would be between 71 and 100%

Answer:

30 cm

Step-by-step explanation:

let x be the lenght of the two sides of equal lenghts, so the other is x+3

and the perimeter is x+x +x +3

P=3x+3

P=3(x+1)

93=3(x+1)

31=x+1

x=30

so the shorter sides are of 30 centimeters and the longest is 33

Answer:

16.50

Step-by-step explanation:

Constant of proportionality = no of gallons of water per 1 minute.

In the first row, we have 16.50 gallons of water per 1 minute.

In the 2nd row, we have 24.75 gallons of water in 1.5 minutes. In 1 minute, we will have 24.75 ÷ 1.5 = 16.50 gallons

In the 3rd row, we have 33 gallons in 2 minutes. In 1 minute, we will have 33 ÷ 2 = 16.50 gallons.

We can see that there seems to be the same constant of proportionality for the 2nd and 3rd row, which is 16.50.

Thus, a relationship between gallons of water (w) and time (t), considering the constant, 16.50, can be written as: [tex] w = 16.50t [/tex]

This means the constant of proportionality, 16.50, is same for all rows.

Answer:

X is uniformly distributed.

Step-by-step explanation:

Uniform Distribution:

This is the type of distribution where all outcome of a certain event have equal likeliness of occurrence.

Example of Uniform Distribution is - tossing a coin. The probability of getting a head is the same as the probability of getting a tail. The have equal likeliness of occurrence.

De acordo com a disponibilidade da unidade, há apenas a seguinte dosagem: 1g/2mL - ou seja, uma grama de dipirona a cada 2mL

O enunciado está meio mal formulado, pois é dito que foram prescritos 500mg de dipirona e é essa quantidade de farmaco que a criança tem que tomar. Deseja-se saber quantos mL deverao ser administrados.

Fazendo a classica regra de 3, podemos chegar no volume desejado:

(atentar que 500mg = 0,5g)

     g               mL

     1    ---------   2

    0,5  ---------  X    

1 . X = 0,5 . 2

Answer:

Simple Interest : $ 4400

Step-by-step explanation:

We want to calculate the interest on $ 10,000, at 5.5% interest rate per year, over a course of 8 years.

We can use the simple interest formula here, or :

I = P × r × t,

Where P is the principle amount, $ 10,000, r is the interest rate, 5.5% each year, or in decimal form 5.5 / 100 = 0.055. t is the time, 8 years.

Simple Interest : 10000 × 0.055 × 8 =  $4400.00

Then again the interest can be added to the principal amount ( $10,000 ) to receive some new amount after 8 years, which is $ 14,000. However the simple interest earned in 8 years at a rate of 5.5% should be $4400.

The simple interest earned on the amount is $4,400

Interest is the total amount that would be paid or earned from making an investment or taking a loan over a period of time.

Simple Interest  = principal x time x interest rate

principal = amount borrowed = $10,000

time = 8 years

Interest rate = 5.5%

10,000 x 0.055 x 8 = $4,400

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Answer:

300.05 miles

Step-by-step explanation:

initial fee= $39.99

final bill = $ 100

cost =$ 0.20 per mile

remaining amount = $ 60.01

solution,

she drive = remaining amount / cost

=60.01/0.20

=300.05 miles

Answer:

500 miles

Step-by-step explanation:

Let us use cross multiplication to find the unknown amount.

Given:

1) Cost for 1 mile=$0.20

2)Cost for x miles=$100

Solution:

No of miles                             Cost

1) 1                                             $0.20

2)x                                             $100

By cross multiplying,

100 x 1= 0.20x

x=100/0.20

x=500 miles

Thank you!

Answer:

[tex]18\sqrt2[/tex]

Step-by-step explanation:

To simplify:

[tex]2 \sqrt{18}+ 3 \sqrt2+ \sqrt{162 }[/tex]

First of all, let us write 18 and 162 as product of prime factors:

[tex]18 = 2 \times \underline{3 \times 3}\\162 = 2 \times \underline{3 \times 3} \times \underline{3 \times 3}[/tex]

The pairs are underlined as above.

While taking roots, only one of the numbers from the pairs will be chosen.

Now, taking square roots.

[tex]\sqrt{18} =3 \sqrt2[/tex]

[tex]162 = 3 \times 3 \times \sqrt 2 = 9 \sqrt2[/tex]

So, the given expression becomes:

[tex]2 \sqrt{18}+ 3 \sqrt2+ \sqrt{162 } = 2 \times 3\sqrt2 + 3\sqrt2 +9\sqrt2\\\Rightarrow 6\sqrt2 + 3\sqrt2 +9\sqrt2\\\Rightarrow \sqrt2(6+3+9)\\\Rightarrow \bold{18\sqrt2}[/tex]

So, the answer is:

[tex]18\sqrt2[/tex] or 18 StartRoot 2 EndRoot

Answer:

its B. 18 sqrt(2)

Step-by-step explanation:

just took test

Answer:

A. 10%

B. Lower limit= 21

Upper limit = 39

Step-by-step explanation:

Mean = 30

SD = 3

a. COV = SD/|x| × 100

= 3/30 × 100

= 10%

= 0.1

B. For 88.9 chevbychev interval:

= (1 - 1/K²) = 0.889

= 1/K² = 1 - 0.889

= 1/K² = 0.111

= K² = 1/0.111

= K² = 9

= K = √9

K = 3

Lower limit = 30 - 3(3)

Lower limit = 21

Upper limit = 30 + 3(3)

Upper limit = 39

Therefore lower limit is 21 and upper limit is 39

Answer:

Step-by-step explanation:

a) Putting 6 into the first (left) machine will give an output of ...

  y = √(6 -5) = √1 = 1

Putting 1 into the second (right) machine will give an output of ...

  y = 1² -6 = -5

This answers the second question, but not the first question.

__

If we put 6 into the right machine first, we get an output of ...

  y = 6² -6 = 30

Putting 30 into the left machine, we get an output of ...

  y = √(30 -5) = √25 = 5 . . . . . the desired output.

The input must go into the right machine first, then its output goes into the left machine to get a final output of 5 from an input of 6.

__

b) The left machine cannot produce negative outputs, so achieving an output of -5 with the arrangement used in part A is not possible. (green curves in the attached graph)

However, as we have shown above, inputting 6 to the left machine first, following that by processing with the right machine, can produce an output of -5. (purple curve in the attached graph)

Answer:

Step-by-step explanation:

To find the approximate value of y when

x = 13 and z = 195 we must first find the relationship between them

The statement

y varies jointly with x and z is written as

where k is the constant of proportionality

From the question

y = 2

x = 11

z = 110

We have

2 = 11(110)k

2 = 1210k

Divide both sides by 1210

[tex]k = \frac{1}{605} [/tex]

So the formula for the variation is

[tex]y = \frac{1}{605} xz[/tex]

When

x = 13

z = 195

y is

[tex]y = \frac{1}{605} (13)(195)[/tex]

[tex]y = \frac{507}{121} [/tex]

y = 4.1900

We have the final answer as

Hope this helps you

Answer:

Below

Step-by-step explanation:

● 8 ÷ (3-x)

Dividing by 3-x is like multiplying by 1/(3-x)

● 8 × (1/3-x)

● 8 /(3-x)

Answer:

A.the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. β  = 0.0122

C. β  = 0.0000

Step-by-step explanation:

Given that:

Mean = 100

standard deviation = 2

sample size = 9

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

[tex]\mathtt{H_o: \mu = 100}[/tex]

[tex]\mathtt{H_1: \mu \neq 100}[/tex]

A. If the acceptance region is defined as [tex]98.5 < \overline x > 101.5[/tex] , find the type I error probability [tex]\alpha[/tex] .

Assuming the critical region lies within [tex]\overline x < 98.5[/tex] or [tex]\overline x > 101.5[/tex], for a type 1 error to take place, then the sample average x will be within the critical region when the true mean heat evolved is [tex]\mu = 100[/tex]

∴

[tex]\mathtt{\alpha = P( type \ 1 \ error ) = P( reject \ H_o)}[/tex]

[tex]\mathtt{\alpha = P( \overline x < 98.5 ) + P( \overline x > 101.5 )}[/tex]

when  [tex]\mu = 100[/tex]

[tex]\mathtt{\alpha = P \begin {pmatrix} \dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} < \dfrac{\overline 98.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} + \begin {pmatrix}P(\dfrac{\overline X - \mu}{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{101.5 - 100}{\dfrac{2}{\sqrt{9}}} \end {pmatrix} }[/tex]

[tex]\mathtt{\alpha = P ( Z < \dfrac{-1.5}{\dfrac{2}{3}} ) + P(Z > \dfrac{1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) + P(Z > 2.25) }[/tex]

[tex]\mathtt{\alpha = P ( Z <-2.25 ) +( 1- P(Z < 2.25) })[/tex]

From the standard normal distribution tables

[tex]\mathtt{\alpha = 0.0122+( 1- 0.9878) })[/tex]

[tex]\mathtt{\alpha = 0.0122+( 0.0122) })[/tex]

[tex]\mathbf{\alpha = 0.0244 }[/tex]

Thus, the type 1 error probability is [tex]\mathbf{\alpha = 0.0244 }[/tex]

B. Find beta for the case where the true mean heat evolved is 103.

The probability of type II error is represented by β. Type II error implies that we fail to reject null hypothesis [tex]\mathtt{H_o}[/tex]

Thus;

β = P( type II error) - P( fail to reject [tex]\mathtt{H_o}[/tex] )

∴

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 103[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -103}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-103}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-4.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-1.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-6.75 \leq Z \leq -2.25) }[/tex]

[tex]\mathtt{\beta = P(z< -2.25) - P(z < -6.75 )}[/tex]

From standard normal distribution table

β  = 0.0122 - 0.0000

β  = 0.0122

C. Find beta for the case where the true mean heat evolved is 105. This value of beta is smaller than the one found in part (b) above. Why?

[tex]\mathtt{\beta = P(98.5 \leq \overline x \leq 101.5) }[/tex]

Given that [tex]\mu = 105[/tex]

[tex]\mathtt{\beta = P( \dfrac{98.5 -105}{\dfrac{2}{\sqrt{9}}} \leq \dfrac{\overline X - \mu}{\dfrac{\sigma}{n}} \leq \dfrac{101.5-105}{\dfrac{2}{\sqrt{9}}}) }[/tex]

[tex]\mathtt{\beta = P( \dfrac{-6.5}{\dfrac{2}{3}} \leq Z \leq \dfrac{-3.5}{\dfrac{2}{3}}) }[/tex]

[tex]\mathtt{\beta = P(-9.75 \leq Z \leq -5.25) }[/tex]

[tex]\mathtt{\beta = P(z< -5.25) - P(z < -9.75 )}[/tex]

From standard normal distribution table

β  = 0.0000 - 0.0000

β  = 0.0000

The reason why the value of beta is smaller here is that since the difference between the value for the true mean and the hypothesized value increases, the probability of type II error decreases.