Ms.Griffin has a class of 18 students. She can spend $19 on each student to buy math supplies for each year. She first buys all of her students calculators, which costs a total of 88.02. After buying the calculators, how much does she have left to spend on each student

Answer:

See Below.

Step-by-step explanation:

We are given that:

[tex]\displaystyle I = I_0 e^{-kt}[/tex]

Where I₀ and k are constants.

And we want to prove that:

[tex]\displaystyle \frac{dI}{dt}+kI=0[/tex]

From the original equation, take the derivative of both sides with respect to t. Hence:

[tex]\displaystyle \frac{d}{dt}\left[I\right] = \frac{d}{dt}\left[I_0e^{-kt}\right][/tex]

Differentiate. Since I₀ is a constant:

[tex]\displaystyle \frac{dI}{dt} = I_0\left(\frac{d}{dt}\left[ e^{-kt}\right]\right)[/tex]

Using the chain rule:

[tex]\displaystyle \frac{dI}{dt} = I_0\left(-ke^{-kt}\right) = -kI_0e^{-kt}[/tex]

We have:

[tex]\displaystyle \frac{dI}{dt}+kI=0[/tex]

Substitute:

[tex]\displaystyle \left(-kI_0e^{-kt}\right) + k\left(I_0e^{-kt}\right) = 0[/tex]

Distribute and simplify:

[tex]\displaystyle -kI_0e^{-kt} + kI_0e^{-kt} = 0 \stackrel{\checkmark}{=}0[/tex]

Hence proven.

Answer:

$1.65

Step-by-step explanation:

or

Answer:

Step-by-step explanation :

Let % increase in administrative secretary be = x

Let % increase in new faculty receive be = y

Administrative secretary salary = 28,000

New faculty receive Salary = 40,000

(8)*(x/100)* (28000) + (7)*(y/100)*(40000) = 5,000

2240x +2800 y = 5,000

224x +280 y = 500

56x +70y = 125

Therefore, x and y should be chosen such that it satisfy the above equation.

Answer:

The center is (0,0) and the radius is 1/4

Step-by-step explanation:

The formula for a circle is

(x-h)^2 + (y-k)^2 = r^2  where (h,k) is the center and r is the radius

16x^2+16y^2=1

Divide  by 16

x^2+y^2=1/16

x^2 + y^2 = (1/4) ^2

(x-0)^2 + (y-0)^2 = (1/4) ^2

The center is (0,0) and the radius is 1/4

9514 1404 393

Answer:

  B. a = plus-or-minus 4

Step-by-step explanation:

  3a² -21 = 27 . . . . . . . given

  3a² = 48 . . . . . . . . . . add 21 to both sides (desired form)

  a² = 16 . . . . . . . . . . .  divide both sides by 3

  a = ±4 . . . . . . . take the square root

Answer:

A. multi-way split.

Step-by-step explanation:

Multi way split consists of internal at decision tree branches. Gini index measures probability of impurity in the random variables chosen. Entropy is measure of uncertainty in the sample selected. Binary splitting is used to speed up numerical evaluation.

Missing from the question

Aletheia's speed is 0.6 miles per hour slower than Elvira's speed.

Answer:

[tex]s_E = 3.0[/tex]

[tex]s_A = 2.4[/tex]

Step-by-step explanation:

Given

[tex]d = 3.2m[/tex] -- distance

[tex]t_E = 1/2[/tex] --- Elvira time

[tex]t_A = 2/3[/tex] --- Aletheia time

[tex]s_E - s_A = 0.6[/tex] --- the relationship between their speeds

Required

Their walking speed

Distance (d) is calculated as:

[tex]d = speed * time[/tex]

For Elvira, we have:

[tex]d_E = s_E * 1/2[/tex]

For Aletheia, we have:

[tex]d_A = s_A * 2/3[/tex]

So, we have:

[tex]d_E + d_A = d[/tex] --- total distance

This gives:

[tex]s_E * 1/2 + s_A * 2/3 = 3.2[/tex]

Recall that:

[tex]s_E - s_A = 0.6[/tex]

Make sE the subject

[tex]s_E = 0.6+s_A[/tex]

Substitute [tex]s_E = 0.6+s_A[/tex] in [tex]s_E * 1/2 + s_A * 2/3 = 3.2[/tex]

[tex](0.6+s_A)* 1/2 + s_A * 2/3 = 3.2[/tex]

[tex]0.3+1/2s_A + 2/3s_A = 3.2[/tex]

Collect like terms

[tex]1/2s_A + 2/3s_A = 3.2-0.3[/tex]

[tex]1/2s_A + 2/3s_A = 2.9[/tex]

Express all as decimal

[tex]0.5s_A + 0.7s_A= 2.9[/tex]

[tex]1.2s_A= 2.9[/tex]

Divide both sides by 1.2

[tex]s_A = 2.4[/tex]

Recall that:

[tex]s_E = 0.6+s_A[/tex]

So, we have:

[tex]s_E = 0.6+2.4[/tex]

[tex]s_E = 3.0[/tex]  

Answer:

432x^13

Step-by-step explanation:

(4x^2)^2 • (3x^3)^3

We know that a^b^c = a^(b*c)

4^2 x^2^2  * 3^3 x^3^3

16 x^4  * 27 x^9

We know that a^b ^ a^c = a^(b+c)

16*27 x^(4+9)

432x^13

Answer:

432x¹³

Step-by-step explanation:

( 4x² ) ² • ( 3x³ ) ²

( 16x²)² • ( 27x³)²

[tex]16 x{}^{2 \times 2} \times 6 {}^{3 \times 3 } \\ 16x {}^{4} \times27 {}^{9} [/tex]

[tex](16 \times 27)x {}^{4 + 9} [/tex]

432x¹³

Answer:

(3x+2) by (2x+1)

Step-by-step explanation:

A cardboard is a rectangle, and has two dimensions. Given a quadratic equation, you should find a way to split it in two.

The easiest way to do so is through factoring. (There are many ways to do this, take a look at the plethora of sources offered on the internet.)

When the expression 6x^2 + 7x + 2 is factored, it is (3x+2)(2x+1). Hence, these are your dimensions.

Let f(x) = ax³ + 9x² + 4x – 10

g(x) = 0⇒x - 3 = 0

⇒ x = 0 + 3

⇒ x = 3

On dividing f(x) by x - 3, it leaves a remainder 5.

Now keeping, f(3) = 5

⇒a(3)³ + 9(3)² + 4(3) - 10 = 5

⇒ a × 27 + 9 × 9 + 4 × 3 - 10 = 5

⇒ 27a + 81 + 12 - 10 = 5

⇒ 81 + 12 - 10 - 5 = 27a

⇒ 81 + 12 - 15 = 27a

⇒ 93 - 15 = 27a

⇒ 78 = 27a

⇒ a = 27/78

⇒ a = 0.3461

The answer is 4 because the frequency is the number of cycles completed in one interval. Typically, the interval given is 2π. Here, you can count the cycles and get 4.

Answer:

a) 0.3056 = 30.56% probability that 3 females and 2 males are selected.

b) 0.0306 = 3.06% probability that all five students selected are males.

c) 0.0131 = 1.31% probability that all five students selected are females.

d) 0.9869 = 98.69% probability that at least one male is selected.

Step-by-step explanation:

The students are chosen from the sample without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

In this question:

15 + 13 = 28 students, which means that [tex]N = 28[/tex]

5 are selected, which means that [tex]n = 5[/tex]

13 females, which means that [tex]k = 13[/tex]

a. 3 females and 2 males are selected?

3 females, so this is P(X = 3).

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 3) = h(3,28,5,13) = \frac{C_{13,3}*C_{15,2}}{C_{28,5}} = 0.3056[/tex]

0.3056 = 30.56% probability that 3 females and 2 males are selected.

b.all five students selected are males?

0 females, so this is P(X = 0).

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 0) = h(0,28,5,13) = \frac{C_{13,0}*C_{15,5}}{C_{28,5}} = 0.0306[/tex]

0.0306 = 3.06% probability that all five students selected are males.

c. all five students selected are females?

This is P(X = 5). So

[tex]P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}[/tex]

[tex]P(X = 5) = h(5,28,5,13) = \frac{C_{13,5}*C_{15,0}}{C_{28,5}} = 0.0131[/tex]

0.0131 = 1.31% probability that all five students selected are females.

d.at least one male is selected?

Less than five females, so:

[tex]P(X < 5) = 1 - P(X = 5) = 1 - 0.0131 = 0.9869[/tex]

0.9869 = 98.69% probability that at least one male is selected.

Answer:

80 Seconds

I dont really want to type the whole thing out, just think about it again, or go to a tutor website, you should be able to get it, you have to use these, multiplication of three numbers, and multiplication and division by factorization of numbers.

Answer:

C

Step-by-step explanation:

Vertical asymptotes are always in the form x = ?

If you look at the dotted line, it lands on 2.  Because it's a vertical line, the asymptote is going to be x = 2

I need the help for this quick app anyone can help

Answer:

h = 6.2 units

Step-by-step explanation:

Given triangle ABC is a right triangle with the measures of the two sides,

BC = [tex]\frac{10}{2}[/tex] = 5 units

AC = 8 units

By applying Pythagoras theorem in the given triangle,

AC² = AB² + BC²

8² = AB² + 5²

AB² = 64 - 25

AB = √39

AB = 6.24 units

AB ≈ 6.2 units

Answer:

See answers below

Step-by-step explanation:

Given the following functions:

r(x) = x - 6

s(x) = 2x²

r(s(x)) = r(2x²)

Replacing x with 2x² in r(x) will give;

r(2x²) = 2x² - 6

r(s(x)) = 2x² - 6

(r-s)(x) = r(x) - s(x)

(r-s)(x) = x - 6 - 2x²

Rearrange

(r-s)(x) = - 2x²+x-6

(r+s)(x) = r(x) + s(x)

(r-s)(x) = x - 6 + 2x²

Rearrange

(r-s)(x) = 2x²+x-6

The question is incomplete. The complete question is :

In a​ poll, 1100 adults in a region were asked about their online vs.​ in-store clothes shopping. One finding was that 43​% of respondents never​ clothes-shop online. Find and interpret a ​95% confidence interval for the proportion of all adults in the region who never​ clothes-shop online.

Solution :

95% confidence interval for p is :

[tex]$\hat p - Z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}} < p < \hat p + Z_{\alpha/2}\times \sqrt{\frac{\hat p(1-\hat p)}{n}}$[/tex]

[tex]$0.43 - 1.96\times \sqrt{\frac{0.43(1-0.43)}{1100}} < p < 0.43 + 1.96\times \sqrt{\frac{0.43(1-0.43)}{1100}}$[/tex]

0.401 < p < 0.459

Therefore, 95% confidence interval is from 0.401 to 0.459

If a is in the second quadrant, then cos(a) < 0 and sin(a) > 0.

Recall the double angle identity for cosine:

cos(2a) = 2 cos²(a) - 1 = 1 - 2 sin²(a)

It follows that

2 cos²(a) - 1 = -4/5   ==>   cos²(a) = 1/10   ==>   cos(a) = -1/√10

1 - 2 sin²(a) = -4/5   ==>   sin²(a) = 9/10   ==>   sin(a) = 3/√10

Then we find

1/cos(a) = sec(a) = -√10

1/sin(a) = csc(a) = √10/3

sin(a)/cos(a) = tan(a) = -3

1/tan(a) = cot(a) = -1/3

Answer:

t = 2

Step-by-step explanation:

5(6t-3)=5t+35

Step 1 distribute the 5 by multiplying 5 to what is inside of the parenthesis

5(6t) - 5(3) = 5t + 35

Outcome: 30t - 15 = 5t + 35

Step 2 add 15 to both sides

30t - 15 + 15 = 5t + 35 + 15

Outcome: 30t = 5t + 50

Step 3 subtract 5t from both sides

30t - 5t = 5t - 5t + 50

Outcome: 25t = 50

Step 4 divide both sides by 25

25t/25 = 50 we're left with t = 2

9514 1404 393

Explanation:

There are several ways you can go at this. Here are a couple. All proofs will start with the given relations being repeated as part of the proof. Here are the next steps.

∠AED ≅ ∠BAE . . . . alternate interior angles are congruent

∠AED +∠AEB = 90° . . . . angle sum theorem

∠BAE +∠AEB = 90° . . . . substitution property of equality

∠CEF ≅ ∠BCE . . . . alternate interior angles are congruent

∠CEF +∠CEB = 90° . . . . angle sum theorem

∠BCE +∠CEB = 90° . . . . substitution property of equality

∠BAE +∠AEB = ∠BCE +∠CEB . . . . substitution property of equality

∠BAE +∠AEB = ∠BCE +∠AEB . . . . substitution property of equality

∠BAE = ∠BCE . . . . addition property of equality

∠ABE = ∠CBE = 90° . . . . BE ⊥ AC

BE ≅ BE . . . . reflexive property of congruence

ΔBEA ≅ ΔBEC . . . . ASA congruence theorem

∠BAE ≅ ∠BCE . . . . CPCTC

Answer:

didn't get the question did u forget to put the sequence ???????

pls write the question fully so that I can help you

Answer:

5 cases

10 additional units

Step-by-step explanation:

Given that :

Total number of units needed = 60 units

Total number of units per case = 14

Hence, the total number of cases required will be :

Number of units needed / number of units per case

Number of cases required = 60 / 14 = 4.285 (this means that 5 cases are required as 4 cases won't be up to 60 units)

With 5 cases, we have exceeded the required units needed :

Additional units will be : (14 * 5) - 60

Additional units = 70 - 60 = 10 units

Answer:

u = 2

Step-by-step explanation:

Since this is a right triangle, we can use trig functions

cos theta = adj side / hypotenuse

cos 45 = sqrt(2) / u

u cos 45 = sqrt(2)

u = sqrt(2) / cos 45

u = sqrt(2) / 1/ sqrt(2)

u = sqrt(2) * sqrt(2)

u =2

u=2

Answer:

Solution given:

Cos 45°=base/hypotenuse

[tex]\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{u}[/tex]

doing crisscrossed multiplication

[tex]\sqrt{2}*\sqrt{2}=1*u[/tex]

u=2

Answer:

I think the answer is 5/6

Step-by-step explanation:

There are three even numbers and two uneven numbers less than four. Therefore, on a standard die, the.probability of Rollin a number that is even or less than for is 5/6.

Answer:

Symmetric with respect to the x-axis

Symmetric with respect to the y-axis

Symmetric with respect to the origin

An

Step-by-step explon: