-6 = u/ -1 +-4 what's u

i bought 3 small tapas dishes and 2 large ones on a visit to café ba-ba-reeba and. spent $14. I went back another day and got 4

satela [25.4K]

The small dish costs $2 each and the large dish costs $4 each,

2(3)+4(2)=6+8=14

2(4)+4(1)=8+4=12

5 0

3 years ago

Find the percent change from $20 to $25

Viktor [21]

Answer:

20%decrease

Step-by-step explanation:

sorry I'm bad at explaining things

5 0

2 years ago

At 7:00 am, the temperature was -4℉. At 9:00 am, it was 8° warmer. What was the temperature at 9:00 am? *

Nana76 [90]

4℉.

What we know about Degrees is that there is a Positive type and a negative type.

(i.e: 30℉ is positive and -30℉ is negative.)

If the temperature was -4℉ at 7AM, then it is negative. If it goes up by an amount that is more than 4 then that negative will go up to a positive temperature. In this case: At 9AM it was 8° warmer.

Warmer is a keyword. If it is warmer by an amount, Negative temperatures will go up to a positive and positive temperature will just go up. If it gets cooler, negative temperatures will go down further and positive temperatures will go down to a negative.

So lets work out this problem with our newfound knowledge.

-4° F at 7AM

8° warmer at 9AM

-4 + 8 = 4.

The temperature was 4° at 9AM.

-Snooky

8 0

2 years ago

A survey collected data on annual credit card charges in seven different categories of expenditures: transportation, groceries,

Andru [333]

Answer:

A. Null and alternative hypothesis:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

B. Yes. At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

P-value = 0.00002

C. As the difference is calculated as (population 1 − population 2), being population 1: groceries and population 2: dinning out, and knowing there is evidence that the true mean difference is positive, we can say that the groceries annual credit card charge is higher than dinning out annual credit card charge.

The point estimate is the sample mean difference d=$840.

The 95% confidence interval for the mean difference between the population means is (490, 1190).

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

Then, the null and alternative hypothesis are:

$H_0: \mu_d=0\\\\H_a:\mu_d\neq 0$

The significance level is 0.05.

The sample has a size n=42.

The sample mean is M=840.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=1123.

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

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{1123}{\sqrt{42}}=173.2827$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{840-0}{173.2827}=\dfrac{840}{173.2827}=4.848$

The degrees of freedom for this sample size are:

$df=n-1=42-1=41$

This test is a two-tailed test, with 41 degrees of freedom and t=4.848, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=2\cdot P(t>4.848)=0.00002$

As the P-value (0.00002) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

At a significance level of 0.05, there is enough evidence to support the claim that there is signficant difference between the population mean credit card charges for groceries and the population mean credit card charges for dining out.

We have to calculate a 95% confidence interval for the mean difference.

The t-value for a 95% confidence interval and 41 degrees of freedom is t=2.02.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=2.02 \cdot 173.283=350$