Help me please I can use it so hook it up !!

A family discount card offers deals for the cinema. The card states:

Rufina [12.5K]

Answer:

a)£21.80

b) 3 trips

Step-by-step explanation:

The cost of an adult is $10.

A child’s ticket is $6.

FAMILY DISCOUNT CARD

- 45% off all adults tickets

- Children save 2/5 off of full price tickets

a) How much will it cost to buy 2 adult tickets and children’s tickets?

(a)

Adults: 2 × £10 = £20

(45 ÷ 100) x £20 = £9

£20 - £9 = £11

Children: 3 × £6 = £18

2 /5 of £18 = £7.20

£18 - £7.20 = £10.80

Total: £11 + £10.80 = £21.80

b) The cost of a family discount card is $40 per year. How many times in the same year would 2 adults and 3 children need to go before you start to savemoney on the cost of the card?

(b) Cost without discount = £20 + £18 = £38b)

No card

1 trip

1 × £38 = £38

2 × £38 = £76

3 × £ 38 = £114

With card

£40 + (1 × £21.80) = £61.80

2 trips 2 × £38 = £76 £40 + (2 × £21.80) = £83.60

3 trips 3 x £38 = £114 £40 + (3 × £21.80) = £105.40

3 trips before the discount card starts being cheaper than regular price

3 0

3 years ago

Find the value of x. Leave your answer in simplest radical form. The chord measures 14, the diameter measures 18. Hint: use pyth

ehidna [41]

Answer:

x = 4√2

Step-by-step explanation:

<u><em>Find the value of x. Leave your answer in simplest radical form. The chord measures 14, the diameter measures 18. Hint: use pythagorean theorem</em></u>

To solve, we will follow the steps below:

using Pythagoras theorem

opposite² + adjacent² = hypotenuse²

14² + y² = 18²

196 + y² = 324

subtract 196 from both-side of the equation

196 -196 + y² = 324-196

y² =128

Take the square root of both-side

√y² = √128

y =√128

y = √64×2

y =√64×√2

y = 8√2

x = half of y

x = y/2

x = 8√2 ÷ 2

x = 4√2

8 0

3 years ago

I need help with my homework I’m in 5th grade and this is the question “Andy makes a model of a new tool that is 0.24 meter long

elena-s [515]

24 meters long. You move the decimal places 2 to the right.

4 0

4 years ago

A manufacturer of metal fasteners expects to ship an average of 1197.00 boxes of fasteners per day. A random sample of 24 days p

lana66690 [7]

Answer:

The p-value obtained is less than the significance level at which the test was performed, hence, we reject the null hypothesis, accept the alternative hypothesis & conclude that there is evidence showing that the average number of boxes shipped per day is different from 1197.00

Step-by-step explanation:

For hypothesis testing, we first clearly state our null and alternative hypothesis.

The null hypothesis is that there is no evidence showing that the average number of boxes shipped per day is different from 1197.00. That is, there is no significant difference between the number of boxes shipped per day and 1197.00

And the alternative hypothesis is that there is evidence showing that the average number of boxes shipped per day is different from 1197.00

Mathematically, the null hypothesis is

H₀: μ₀ = 1197.00

The alternative hypothesis is

Hₐ: μ₀ ≠ 1197.00

To do this test, we will use the t-distribution because no information on the population standard deviation is known

So, we compute the t-test statistic

t = (x - μ₀)/σₓ

x = sample mean = 1193.69 boxes per day

μ₀ = standard that we're comparing the sample mean with = 1197.00

σₓ = standard error of the sample mean

= (σ/√n)

where n = Sample size = 24 days

σ = standard deviation of the sample = 3.3166 boxes per day

σₓ = (3.3166/√24) = 0.677

t = (1193.69 - 1197) ÷ 0.677

t = -4.89

checking the tables for the p-value of this t-statistic

Degree of freedom = df = n - 1 = 24 - 1 = 23

Significance level = 0.05

p-value (for t = -4.89, at 0.05 significance level, with a two tailed condition) = 0.000061

The interpretation of p-values is that

When the (p-value > significance level), we fail to reject the null hypothesis and when the (p-value < significance level), we reject the null hypothesis and accept the alternative hypothesis.

So, for this question, significance level = 5% = 0.05

p-value = 0.000061

0.000061 < 0.05

Hence,

p-value < significance level

This means that we reject the null hypothesis, accept the alternative hypothesis & conclude that there is evidence showing that the average number of boxes shipped per day is different from 1197.00

Hope this Helps!!!

6 0

3 years ago

At 2 PM, a thermometer reading 80°F is taken outside where the air temperature is 20°F. At 2:03 PM, the temperature reading yiel

Alexeev081 [22]

Use Law of Cooling:
$T(t) = (T_0 - T_A)e^{-kt} +T_A$
T0 = initial temperature, TA = ambient or final temperature

First solve for k using given info, T(3) = 42
$42 = (80-20)e^{-3k} +20 \\ \\ e^{-3k} = \frac{22}{60}=\frac{11}{30} \\ \\ k = -\frac{1}{3} ln (\frac{11}{30})$

Substituting k back into cooling equation gives:
$T(t) = 60(\frac{11}{30})^{t/3} + 20$

At some time "t", it is brought back inside at temperature "x".
We know that temperature goes back up to 71 at 2:10 so the time it is inside is 10-t, where t is time that it had been outside.
The new cooling equation for when its back inside is:
$T(t) = (x-80)(\frac{11}{30})^{t/3} + 80 \\ \\ 71 = (x-80)(\frac{11}{30})^{\frac{10-t}{3}} + 80$
Solve for x:
$x = -9(\frac{11}{30})^{\frac{t-10}{3}} + 80$
Sub back into original cooling equation, x = T(t)
$-9(\frac{11}{30})^{\frac{t-10}{3}} + 80 = 60 (\frac{11}{30})^{t/3} +20$
Solve for t:
$60 (\frac{11}{30})^{t/3} +9(\frac{11}{30})^{\frac{-10}{3}}(\frac{11}{30})^{t/3} = 60 \\ \\ (\frac{11}{30})^{t/3} = \frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}} \\ \\ \\ t = 3(\frac{ln(\frac{60}{60+9(\frac{11}{30})^{\frac{-10}{3}}})}{ln (\frac{11}{30})}) \\ \\ \\ t = 4.959$

This means the exact time it was brought indoors was about 2.5 seconds before 2:05 PM

8 0

4 years ago