Ask question

Ask question

All categories

2 years ago

8

the temperature was 78°f at 3pm. each hour for the next ғoυr hours, the temperature decreased by 3°f. what was the temperature a

t 7 pm?

2 answers:

Bad White [126]2 years ago

4 0

At 7 pm, the temperature was 66*F

Temp = 78 - 3x where x = number of hours
78 - 3(7-3)
78 - 12 = 66

Inga [223]2 years ago

4 0

Use x in an equation

x=66

You might be interested in

Mademuasel [1]

no solution.

The two lines are parallel

(same slope but different y intercepts)

6 0

2 years ago

tankabanditka [31]

Answer:

2520

Step-by-step explanation:

LCM(4, 7, 8, 9, 10, 12) = 2520

5 0

2 years ago

Read 2 more answers

Find the value of X?

Molodets [167]

Answer:

x = 58

Step-by-step explanation:

The exterior angle is equal to the sum of the opposite interior angles

90 = 32+x

Subtract 32 from each side

90-32 = x

58 =x

5 0

3 years ago

The difference of 2 and the<br> product of 3 and k

MA_775_DIABLO [31]

Answer:

3k-2

Step-by-step explanation:

Difference obiously means subtraction in this situation and you are subtraction 2 from the end result. " The product of 3 and k" is just refering to multiplying an unknown value by 3. This unknown value then becomes a variable and 3 becomes the coefficient. When these numbers are put together, they are suggesting multiplication. It can also be written to be less confusing. (3 x k)-2 which is the same thing as 3k-2.

8 0

2 years ago

(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

2 years ago