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3 years ago

If a recipe calls fo 2 tbsp how many tsp is that

Mathematics

1 answer:

wlad13 [49]3 years ago

3 0

6 because each tablespoon is 3 teaspoons. Hope this helps.

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A perfume company claims that the mean weight of ther new perfume is at least 8.9 fluid oz. Express the null hypothesis and the

zlopas [31]

Answer:

$H0: mu = 8.9 fl oz.\\\\$

$Ha: mu ≠8.9 fluid oz$

Step-by-step explanation:

Given that A perfume company claims that the mean weight of ther new perfume is at least 8.9 fluid oz

For testing this claim, in Statistics we perform a certain measures called hypothesis testing.

For this first step is to create null and alternate hypothesis.

Normally null hypothesis would have some statistic = something

Here we want to test the mean weight of perfume

Hence null hypothesis would be

H0: mu = 8.9 fl oz.

Alternate hypothesis would be opposite of this claim

i.e.

Ha: mu ≠8.9 fluid oz

Hence answer is

$H0: mu = 8.9 fl oz.\\\\$

$Ha: mu ≠8.9 fluid oz$

6 0

3 years ago

Mrs. DeMarco wants to estimate the length of her porch so she knows how much paint to buy. What is the best benchmark for her to

Zigmanuir [339]

The answer c ,I hope I’m right

8 0

3 years ago

There are 50 read beans and 50 white beans in a bag . The first bean selected is red . The red bean is returned to the bag . Wha

andrew11 [14]

Answer:

50% as there is 100 beans all together

4 0

3 years ago

Find the scale factor using the distance formula.

Julli [10]

1/2 because everything lose half its value points wise.

3 0

2 years ago

I’m Really lost if I could get a answer it would be greatly appreciated

Nataly [62]

<h3>Answers:</h3>

$f(g(x)) = \sqrt{x^2+5}+5\\\\g(f(x)) = x+30+10\sqrt{x-1}$

================================================

Work Shown:

Part 1

$f(x) = \sqrt{x-1}+5\\\\f(g(x)) = \sqrt{g(x)-1}+5\\\\f(g(x)) = \sqrt{x^2+6-1}+5\\\\f(g(x)) = \sqrt{x^2+5}+5\\\\$

Notice how I replaced every x with g(x) in step 2. Then I plugged in g(x) = x^2+6 and simplified.

------------------

Part 2

$g(x) = x^2+6\\\\g(f(x)) = \left(f(x)\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}+5\right)^2+6\\\\g(f(x)) = \left(\sqrt{x-1}\right)^2+2*5*\sqrt{x-1}+\left(5\right)^2+6\\\\g(f(x)) = x-1+10\sqrt{x-1}+25+6\\\\g(f(x)) = x+30+10\sqrt{x-1}\\\\$

In step 4, I used the rule (a+b)^2 = a^2+2ab+b^2

In this case, a = sqrt(x-1) and b = 5.

You could also use the box method as a visual way to expand out $\left(\sqrt{x-1}+5\right)^2$

6 0

3 years ago