Does anyone know ill give brainliest answer if you know

What’s this help pls

lys-0071 [83]

How to solve your problem

9−4+3

9x−4+3x9x-4+3x9x−4+3x

Simplify

1

Combine like terms

9−4+3

9x−4+3x{\color{#c92786}{9x}}-4+{\color{#c92786}{3x}}9x−4+3x

12−4

12x−4{\color{#c92786}{12x}}-412x−4

Solution

12−4

SO D

7 0

3 years ago

Read 2 more answers

A new cookie cake company is offering a

gtnhenbr [62]

Answer:

200.96in^2

Step-by-step explanation:

If the 1st cookie cake is 8" for its diameter and the 2nd cookie cake will have twice the length diameter of the 1st cookie cake .You will do 2 × 8 to get the diameter of the 2nd cookie cake.

First cookie cake diameter is 8".

Second cookie cake's diameter is 16".

Since we need the radius to find the area of the 2nd cookie cake we're going to do 16 / 2 which equals 8 so 8 is the radius.

A=(pi)r^2

A=(pi)8^2 *to the 2nd power means times by the same number

A=(pi)64

A= (3.14)(64)

A=200.96

=200.96^2

pi= 3.14 *for this problem

7 0

3 years ago

Determine the equation for the given line in

Tema [17]

Answer:

Y= -3/5 - 1

Step-by-step explanation:

if you start on the y intercept an go down once (-1)

then go down three times (-3)

positive 5 so go right 5 times it lands on the point :)

im not a teacher and i took this last year so sorry if my explanation isn't the best

6 0

3 years ago

Read 2 more answers

(A) A small business ships homemade candies to anywhere in the world. Suppose a random sample of 16 orders is selected and each

Leviafan [203]

Following are the solution parts for the given question:

For question A:

$\to (n) = 16$

$\to (\bar{X}) = 410$

$\to (\sigma) = 40$

In the given question, we calculate $90\%$ of the confidence interval for the mean weight of shipped homemade candies that can be calculated as follows:

$\to \bar{X} \pm t_{\frac{\alpha}{2}} \times \frac{S}{\sqrt{n}}$

$\to C.I= 0.90\\\\\to (\alpha) = 1 - 0.90 = 0.10\\\\ \to \frac{\alpha}{2} = \frac{0.10}{2} = 0.05\\\\ \to (df) = n-1 = 16-1 = 15\\\\$

Using the t table we calculate $t_{ \frac{\alpha}{2}} = 1.753$ When $90\%$ of the confidence interval:

$\to 410 \pm 1.753 \times \frac{40}{\sqrt{16}}\\\\ \to 410 \pm 17.53\\\\ \to392.47 < \mu < 427.53$

So $90\%$ confidence interval for the mean weight of shipped homemade candies is between $392.47\ \ and\ \ 427.53$ .

For question B:

$\to (n) = 500$

$\to (X) = 155$

$\to (p') = \frac{X}{n} = \frac{155}{500} = 0.31$

Here we need to calculate $90\%$ confidence interval for the true proportion of all college students who own a car which can be calculated as

$\to p' \pm Z_{\frac{\alpha}{2}} \times \sqrt{\frac{p'(1-p')}{n}}$

$\to C.I= 0.90$

$\to (\alpha) = 0.10$

$\to \frac{\alpha}{2} = 0.05$

Using the Z-table we found $Z_{\frac{\alpha}{2}} = 1.645$

therefore $90\%$ the confidence interval for the genuine proportion of college students who possess a car is

$\to 0.31 \pm 1.645\times \sqrt{\frac{0.31\times (1-0.31)}{500}}\\\\ \to 0.31 \pm 0.034\\\\ \to 0.276 < p < 0.344$

So $90\%$ the confidence interval for the genuine proportion of college students who possess a car is between $0.28 \ and\ 0.34.$

For question C:

In question A, We are $90\%$ certain that the weight of supplied homemade candies is between 392.47 grams and 427.53 grams.
In question B, We are $90\%$ positive that the true percentage of college students who possess a car is between 0.28 and 0.34.

Learn more about confidence intervals:

brainly.in/question/16329412

7 0

3 years ago

Suppose a college student pays $500 for tuition fees. However, she also has to pay $250 for her textbooks (ouch!). What percent

Dominik [7]

750 dollars in total <=500+250=750
250/750=1/3 or 33.33333333%

7 0

3 years ago