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3 years ago

Please help I’m so confused 25 points and brainliest!!

Mathematics

1 answer:

Soloha48 [4]3 years ago

4 0

Answer:

This should be correct ( I havent done it in 3 years but it seems right)

Step-by-step explanation:

1.c≈15.71

2.c≈31.42

3.d=12.5

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I know the answer to number 7 but I just want to see if you guys know it

elixir [45]

Plugging 10 for k in the expression gives us
$2(10)^2+5(10)+1$ We simplify this expression to $200+50+1=251$
251 is your answer

4 0

3 years ago

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Marc baked brownies in the brownie pan shown below. Each day, he is going to eat 20 square centimeters of brownie from the pan.

podryga [215]

Answer:

$35\ days$

Step-by-step explanation:

step 1

Find the area of the trapezoid (brownie pan)

$A=\frac{1}{2}(29+41)20=700\ cm^{2}$

step 2

To find the number of days , divide the area of the brownie pan by 20 square centimeters of brownie

$\frac{700}{20}=35\ days$

8 0

3 years ago

WILL GIVE BRAINLIEST& 15 PTS.

kicyunya [14]

Answer:

The given fraction $\frac{x^3-x^2}{x^3}$ reduces to $\frac{x-1}{x}$

Step-by-step explanation:

Consider the given fraction $\frac{x^3-x^2}{x^3}$

We have to reduce the fraction to the lowest terms.

Consider numerator $x^3-x^2$

We can take x² common from both the term,

Thus, numerator can be written as $x^2(x-1)$

Given expression can be rewritten as ,

$\frac{x^3-x^2}{x^3}=\frac{x^2(x-1)}{x^3}$

We can now cancel $x^2$ from both numerator and denominator,

$\Rightarrow \frac{x^2(x-1)}{x^3}=\frac{x^2(x-1)}{x^2 \cdot x}$

$\Rightarrow \frac{(x-1)}{x^2 \cdot x}=\frac{x-1}{x}$

Thus, the given fraction $\frac{x^3-x^2}{x^3}$ reduces to $\frac{x-1}{x}$

6 0

3 years ago

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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:

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