PLEASE HELP ME ASAP!!!!! WILL GIVE BRAINLIEST TO FIRST CORRECT ANSWER!!!!!

Im doing my final. Answer and you get whatever you want.

irina1246 [14]

Answer:it will be a slight probability

Step-by-step explanation:

3 0

3 years ago

a bag contains 14 red tiles, 20 blue tiles, and 16 green tiles, a tile is randomly drawn from the bag. if this is done 40 times

ololo11 [35]

Total=14+20+16=50
P(blue)=20/50=2/5

so, answer=(2/5)*40=16

7 0

3 years ago

Read 2 more answers

Break apart addendsas tens and ones 33 +49

Serhud [2]

Answer:

There are 8 tens and 2 ones

Step-by-step explanation:

Hope this helps

5 0

3 years ago

A random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specime

MrRissso [65]

Answer:

We conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

We conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

Step-by-step explanation:

We are given a random sample of soil specimens was obtained, and the amount of organic matter (%) in the soil was determined for each specimen;

1.10, 5.09, 0.97, 1.59, 4.60, 0.32, 0.55, 1.45, 0.14, 4.47, 1.20, 3.50, 5.02, 4.67, 5.22, 2.69, 3.98, 3.17, 3.03, 2.21, 0.69, 4.47, 3.31, 1.17, 0.76, 1.17, 1.57, 2.62, 1.66, 2.05.

Let $\mu$ = true average percentage of organic matter

So, Null Hypothesis, $H_0$ : $\mu$ = 3% {means that the true average percentage of organic matter in such soil is 3%}

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is something other than 3%}

The test statistics that will be used here is One-sample t-test statistics because we don't know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean percentage of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, the test statistics = $\frac{2.481-3}{\frac{1.616}{\sqrt{30} } }$ ~ $t_2_9$

= -1.76

The value of t-test statistics is -1.76.

(a) Now, at 10% level of significance the t table gives a critical value of -1.699 and 1.699 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics doesn't lie within the range of critical values of t, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is something other than 3% at 10% significance level.

(b) Now, at 5% level of significance the t table gives a critical value of -2.045 and 2.045 at 29 degrees of freedom for the two-tailed test.

Since the value of our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region.

Therefore, we conclude that the true average percentage of organic matter in such soil is 3% at 5% significance level.

8 0

3 years ago

Evaluate the expression. 38 +16 - 12 - 2 - (302) Enter your answer in the box.

masha68 [24]

Answer:

-262

Step-by-step explanation:

Evaluate Subtraction & Addition from right to left [or left to right, as long as you know what you are doing]:

38 + 16 - 12 - 2 - 302

|_______||____________|

$44 - 316 = -262$

I am joyous to assist you anytime.

5 0

3 years ago