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3 years ago

What are the Missing numbers in this sequence: 24, _, 28, 13, 33, 15, 39, _, _.

Mathematics

2 answers:

Yuliya22 [10]3 years ago

8 0

Answer:

12, 18, 46

Step-by-step explanation:

count the gap between 24 and 28. then 28 and 33. and 33 and 39. it starts at going up by 4, then 5, then 6. following this pattern, you would count up 7 from 39 to get your next number

it's the same thing with 13, and 15. the gap between them is 2. that means the next one would be a gap of three, resulting in 18. that also means the one behind it would be a gap of 1, making the missing number 12.

hope this helps,

noah

Free_Kalibri [48]3 years ago

5 0

12,18,46 are the missing numbers

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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

A study claims that college students spend an average of 4 hours or less studying per day. A researcher wants to check if this c

kirill115 [55]

<h2>Answer with explanation:</h2>

Let $\mu$ be the population mean.

By observing the given information, we have :-

$H_0:\mu\leq4\\\\H_a:\mu>4$

Since the alternative hypotheses is left tailed so the test is a right-tailed test.

We assume that the time spend by students per day is normally distributed.

Given : Sample size : n=121 , since n>30 so we use z-test.

Sample mean : $\overline{x}=3.15$

Standard deviation : $\sigma=1.2$

Test statistic for population mean :-

$z=\dfrac{\overline{x}-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

$\Rightarrow\ z=\dfrac{3.15-4}{\dfrac{1.2}{\sqrt{121}}}\\\\\Rightarrow\ z=-7.79166666667\approx-7.79$

Critical value (one-tailed) corresponds to the given significance level :-

$z_{\alpha}=z_{0.1}=1.2816$

Since the observed value of z (-7.79) is less than the critical value (1.2816) , so we do not reject the null hypothesis.

Hence, we conclude that we have enough evidence to accept that the college students spend an average of 4 hours or less studying per day.

4 0

3 years ago

Can someone help me find the area

lisabon 2012 [21]

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Answer: 116.12

Explanation:

13 x 7 = 91

13 - 5 = 8

8 / 2 = 4

3.14(4)^2 = 50.24

50.24 / 2 = 25.12

91 + 25.12 = 116.12

I hope this helped!

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5 0

3 years ago

Read 2 more answers

Need help with my homework

Volgvan

Answer:

$\displaystyle y=\frac{16-9x^3}{2x^3 - 3}$

$\displaystyle y=-\frac{56}{13}$

Step-by-step explanation:

<u>Equation Solving</u>

We are given the equation:

$\displaystyle x=\sqrt[3]{\frac{3y+16}{2y+9}}$

To make y as a subject, we need to isolate y, that is, leaving it alone in the left side of the equation, and an expression with no y's to the right side.

We have to make it in steps like follows.

Cube both sides:

$\displaystyle x^3=\left(\sqrt[3]{\frac{3y+16}{2y+9}}\right)^3$