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3 years ago

PLS HELP BEST ANSWER GETS BRAINLIEST

Mathematics

1 answer:

Blizzard [7]3 years ago

4 0

Answer: 4 cakes
Explanation: 2/3 is equal to 4/6. Each cake only takes 1/6 lbs. of sugar

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Given f ( x ) = 1 x + 10 , find the average rate of change of f ( x ) on the interval [ 8 , 8 + h ] . Your answer will be an exp

son4ous [18]

Answer: 1

Step-by-step explanation:

$f(8+h)=8+h+10=18+h\\\\f(8)=8+10=18$

So, the average rate of change is:

$\frac{(18+h)-18}{(8+h)-8}=\frac{h}{h}=\boxed{1}$

The question is wrong, the expression will not be in terms of h since it is a linear function.

6 0

2 years ago

HELP PLEASE 5/12-(x-3)/6≤(x-2)/3 Solve with a step by step solution

zlopas [31]

Answer:x>3 1/6

Step-by-step explanation:

5/12-(x-3)/6 <(x-2)/3

5/12 *12-(x-3)/6*12<(x-2)/6*12

5-2(x-3)<4(x-2)

5-2x+6<4x-8

-2x+11<4x-8

-2x<4x-19

-6x<-19

-6x(-1)<-19(-1)

6x>19

x>19/6

x>3 1/6

just change all the inequality signs to a x is greater or equal to 3 1/6 and on the rest change to and < and equal sign only put > or equal sign on the last three lines

8 0

3 years ago

What is the range of the function y=3x+8?

jolli1 [7]

Translation, alternates between

7 0

3 years ago

Read 2 more answers

Determine whether each set of data represents a linear, an exponential, or a quadratic function. (Desmos)

Cerrena [4.2K]

Answer:

Linear: The first and third one. (0,5) & (1,1)

Exponential: The second one (above). (-2, 1/16)

Quadratic function: The last one (below). (-3,35)

Step-by-step explanation:

4 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

Shalnov [3]

Answer:

We conclude that the true average percentage of organic matter in such soil is different from 3%.

Step-by-step explanation:

We are given that the values of the sample mean and sample standard deviation are 2.481 and 1.616, respectively.

Suppose we know the population distribution is normal, we have to test the hypothesis that does this data suggest that the true average percentage of organic matter in such soil is something other than 3%.

Let $\mu$ = true average percentage of organic matter in such soil

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

Alternate Hypothesis, $H_A$ : $\mu \neq$ 3% {means that the true average percentage of organic matter in such soil is different 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 amount of organic matter = 2.481%

s = sample standard deviation = 1.616%

n = sample of soil specimens = 30

So, test statistics = $\frac{0.02481-0.03}{{\frac{0.01616}{\sqrt{30} } } }$ ~ $t_2_9$

= -1.759

Now, P-value of the test statistics is given by;

P-value = P( $t_2_9$ > -1.759) = 0.046 or 4.6%

If the P-value of test statistics is more than the level of significance, then we will not reject our null hypothesis as it will not fall in the rejection region.

If the P-value of test statistics is less than the level of significance, then we will reject our null hypothesis as it will fall in the rejection region.

Now, here the P-value is 0.046 which is clearly smaller than the level of significance of 0.05 (for two-tailed test), so we will 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 different from 3%.

3 0

3 years ago