All categories

3 years ago

Hellooo menejkfnfkdjdkdkdmmdmfjf

Mathematics

1 answer:

Vesna [10]3 years ago

6 0

Helloo, i don’t see the image seen above, is that the only picture?

You might be interested in

Solve the equation.<br> -0.3x = - 1.5

frutty [35]

The answer is easy it is x=5

8 0

3 years ago

Read 2 more answers

The tail of a kite is 1.5 feet plus twice the length of the kite. Together, the kite and tail are 15.5 ft long. Find the length

Dimas [21]

Answer:

The answer would be D because 1.5 plus twice the length of the kite would be 2k. 1.5+2k

Hope this helps

4 0

3 years ago

Read 2 more answers

Increase 56 in ratio 2:3

Galina-37 [17]

Answer:37 1/3

Step-by-step explanation:

if you multiply 56 (2/3)=37 1/3

3 0

2 years ago

Read 2 more answers

8x-3(-z-y)<br> X= 1 <br> Y=3 <br> Z=-2

Masteriza [31]

Answer:

Step-by-step explanation:

8x - 3(-z - y)

Substitute

8(1) - 3(-(-2) - 3)

Simplify

8 - 3(-1)

Multiply

8 + 3

Add

8 0

3 years ago

Read 2 more answers

The caffeine content (in mg) was examined for a random sample of 50 cups of black coffee dispensed by a new machine. The mean an

Genrish500 [490]

Answer:

We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 mg .

Step-by-step explanation:

Given -

The sample size is large then we can use central limit theorem

n = 50 ,

Standard deviation $(\sigma)$ = 7.1

Mean $\overline{(y)}$ = 110

$\alpha =$ 1 - confidence interval = 1 - .98 = .02

$z_{\frac{\alpha}{2}}$ = 2.33

98% confidence interval for the mean caffeine content for cups dispensed by the machine = $\overline{(y)}\pm z_{\frac{\alpha}{2}}\frac{\sigma}\sqrt{n}$

= $110\pm z_{.01}\frac{7.1}\sqrt{50}$

= $110\pm 2.33\frac{7.1}\sqrt{50}$

First we take + sign

$110 + 2.33\frac{7.1}\sqrt{50}$ = 112.34

now we take - sign

$110 - 2.33\frac{7.1}\sqrt{50}$ = 107.66

We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 .

5 0

3 years ago