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3 years ago

PLEASE HELP ME!!!

Mathematics

2 answers:

MArishka [77]3 years ago

7 0

Rent answer A you’re welcome

Llana [10]3 years ago

3 0

Answer:

Rent

Step-by-step explanation:

Yes

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If f(x) = x2 – 2x + 3, then find<br> f(-3).

Georgia [21]

Answer:

f(-3) = 18

Step-by-step explanation:

We need just substitute x with (-3) and calculate it.

f(-3) = (-3)² -2*(-3) + 3 = 9 + 6 +3 = 18

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3 years ago

On the day a coffee shop first opened it had 140 customers 65% of the customers ordered a large coffee how manny customers order

madam [21]

91 patrons of the coffee shop ordered a large

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3 years ago

Read 2 more answers

The school that Scott goes to is selling tickets to a fall musical. On the first day of ticket sales the school sold 12 adult ti

harina [27]

Answer:

The price of one adult ticket = $13

The price of one student ticket = $4

Step-by-step explanation:

Let the price of 1 adult ticket = x

Let the price of 1 student ticket = y

On the first day of ticket sales the school sold 12 adult tickets and 10 student tickets for a total of $196.

12 × x + 10 × y = $196

12x + 10y = 196....... Equation 1

The school took in $59 on the second day by selling 3 adult tickets and 5 student tickets

3 × x + 5 × y = $59

3x + 5y = 59.......... Equation 2

Using Elimination method

We eliminate y, by Multiplying equation 1 by 5 and Equation 2 by 10

12x + 10y = 196....... Equation 1 × 5

3x + 5y = 59.......... Equation 2 × 10

60x + 50y = 980....... Equation 3

30x + 50y = 590......... Equation 4

Subtract Equation 4 from Equation 3

30x = 390

x = 390/30

x = $13

Therefore, the price of one adult ticket = $13

Remember: x = $13

3x + 5y = 59.......... Equation 2

Substitute

3(13) + 5y = 59

39 + 5y = 59

5y = 59 - 39

5y = 20

y = 20/5

y = $4

Therefore, the price of one student ticket = $4

8 0

3 years ago

A bag contains 5 milk and 9 dark chocolates.

-Dominant- [34]

Answer:

5/9

Step-by-step explanation:

5 0

2 years ago

A student carried out an experiment to determine the amount of vitamin C in a tablet sample. He performed 5 trials to produce th

ivolga24 [154]

Answer:

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

P-value = 0.166.

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{5}(490+502+505+495+492)\\\\\\M=\dfrac{2484}{5}\\\\\\M=496.8\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{4}((490-496.8)^2+(502-496.8)^2+(505-496.8)^2+(495-496.8)^2+(492-496.8)^2)}\\\\\\s=\sqrt{\dfrac{166.8}{4}}\\\\\\s=\sqrt{41.7}=6.5\\\\\\$

Then, we can perform the hypothesis t-test for the mean.

The claim is that the amount of vitamin C in a tablet sample is different from 500 mg.

Then, the null and alternative hypothesis are:

$H_0: \mu=500\\\\H_a:\mu< 500$

The significance level is 0.05.

The sample has a size n=5.

The sample mean is M=496.8.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.5.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{6.5}{\sqrt{5}}=2.907$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{496.8-500}{2.907}=\dfrac{-3.2}{2.907}=-1.1$

The degrees of freedom for this sample size are:

$df=n-1=5-1=4$

This test is a left-tailed test, with 4 degrees of freedom and t=-1.1, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.166) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the amount of vitamin C in a tablet sample is different from 500 mg.

4 0

3 years ago