I need help I'm not good in math

A teacher has been given £35 to spend on pencils and rulers.

SSSSS [86.1K]

Packet of 12 pencils = p
Packet of 30 rulers = r

p = £2.15
r = £6.00

£35 = 4r + xp
£35 = 4(£6.00) + x(£2.15)
£35 = £24 + £2.15x
£11 = £2.15x
x = 5.12

Since she can buy 5.12 packets of pencils, it is equal to 5 packets of pencils.

3 0

2 years ago

Calculate the data value that corresponds to each of the following z-scores.

Fynjy0 [20]

Answer:

a) 90.1

b) $35.6

c) 3.0915 hours

Step-by-step explanation:

The z-score measures how many standard deviations a score X is above or below the mean.

It is given by the following formula:

$Z = \frac{X - \mu}{\sigma}$

In which $\mu$ is the mean and $\sigma$ is the standard deviaition.

In all three cases, we have to find X

a. Final exam scores: Allison’s z-score = 2.30, μ = 74, σ = 7.

$Z = \frac{X - \mu}{\sigma}$

$2.30 = \frac{X - 74}{7}$

$X - 74 = 7*2.3$

$X = 90.1$

b. Weekly grocery bill: James’ z-score = –1.45, μ = $53, σ = $12.

$Z = \frac{X - \mu}{\sigma}$

$-1.45 = \frac{X - 53}{12}$

$X - 53 = -1.45*12$

$X = 35.6$

Mean and standard deviation in dollars, so the answer also in dollars.

c. Daily video game play time: Eric’s z-score = –0.79, μ = 4.00 hours, σ = 1.15 hours.

$Z = \frac{X - \mu}{\sigma}$

$-0.79 = \frac{X - 4}{1.15}$

$X - 4 = -0.79*1.15$

$X = 3.0915$

Mean and standard deviation in hours, answer in hours.

6 0

3 years ago

A right cylinder has the radius of 3.5cm and the height of 11cm if i place 3 ice cube with edge length of 2cm in the glass what

Brrunno [24]

Answer:

Volume of cylinder =

$\pi \times r {}^{2} \times h$

$\pi \times 3.5 {}^{2} \times 11 = 423cm {}^{3}$

Volume of 1 ice cube = 2*2*2= 8cm3

Volume of 3 ice cubes= 3*8= 24cm3

Volume of juice added = 423-24 = 399 cm3

4 0

3 years ago

The indicated function y1(x is a solution of the given differential equation. use reduction of order or formula (5 in section 4.

Taya2010 [7]

Given a solution $y_1(x)=\ln x$ , we can attempt to find a solution of the form $y_2(x)=v(x)y_1(x)$ . We have derivatives

$y_2=v\ln x$
${y_2}'=v'\ln x+\dfrac vx$
${y_2}''=v''\ln x+\dfrac{v'}x+\dfrac{v'x-v}{x^2}=v''\ln x+\dfrac{2v'}x-\dfrac v{x^2}$

Substituting into the ODE, we get

$v''x\ln x+2v'-\dfrac vx+v'\ln x+\dfrac vx=0$
$v''x\ln x+(2+\ln x)v'=0$

Setting $w=v'$ , we end up with the linear ODE

$w'x\ln x+(2+\ln x)w=0$

Multiplying both sides by $\ln x$ , we have

$w' x(\ln x)^2+(2\ln x+(\ln x)^2)w=0$

and noting that

$\dfrac{\mathrm d}{\mathrm dx}\left[x(\ln x)^2\right]=(\ln x)^2+\dfrac{2x\ln x}x=(\ln x)^2+2\ln x$

we can write the ODE as

$\dfrac{\mathrm d}{\mathrm dx}\left[wx(\ln x)^2\right]=0$

Integrating both sides with respect to $x$ , we get

$wx(\ln x)^2=C_1$
$w=\dfrac{C_1}{x(\ln x)^2}$

Now solve for $v$ :

$v'=\dfrac{C_1}{x(\ln x)^2}$
$v=-\dfrac{C_1}{\ln x}+C_2$

So you have

$y_2=v\ln x=-C_1+C_2\ln x$

and given that $y_1=\ln x$ , the second term in $y_2$ is already taken into account in the solution set, which means that $y_2=1$ , i.e. any constant solution is in the solution set.

4 0

3 years ago

Answer? BRAINLIEST IF CORRECT

alexira [117]

The answer is 12 your welcome

4 0

3 years ago