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Vlad1618 [11]
3 years ago
7

A recipe calls for 2 cups of milk for each batch of pancakes. Jamie would like to triple her recipe. How many cups of milk will

she need?
Mathematics
1 answer:
Arturiano [62]3 years ago
5 0
Triple means 3 times so 2 x 3 =6
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2/3=1.2/x solve for x
Anvisha [2.4K]

First isolate x by multiplying by x on both sides.  It will look like this:

(2/3)x=1.2

Then devise by (2/3) on both sides to get:

x= 1.2/(2/3)

The easiest way to get this is plug that into a calculator and get the answer:

x=1.8

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Find the area of the figure by subtraction.
arsen [322]

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<em>56</em>

Step-by-step explanation:

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2 years ago
The mean MCAT score 29.5. Suppose that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.
Paul [167]

Answer:

We conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

Step-by-step explanation:

We are given that the Kaplan tutoring company obtains a sample of 40 students with a mean MCAT score of 32.2 with a standard deviation of 4.2.

Let \mu = <u><em>population mean score</em></u>

So, Null Hypothesis, H_0 : \mu \leq 29.5      {means that the students that took the Kaplan tutoring have a mean score less than or equal to 29.5}

Alternate Hypothesis, H_A : \mu > 29.5      {means that the students that took the Kaplan tutoring have a mean score greater than 29.5}

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

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

where, \bar X = sample mean MCAT score = 32.2

            s = sample standard deviation = 4.2

            n = sample of students = 40

So, <u><em>the test statistics</em></u> =  \frac{32.2-29.5}{\frac{4.2}{\sqrt{40} } }  ~  t_3_9

                                    =  4.066

The value of t-test statistics is 4.066.

Now, at 0.05 level of significance, the t table gives a critical value of 1.685 at 39 degrees of freedom for the right-tailed test.

Since the value of our test statistics is more than the critical value of t as 4.066 > 1.685, so <u><em>we have sufficient evidence to reject our null hypothesis</em></u> as it will fall in the rejection region.

Therefore, we conclude that the students that took the Kaplan tutoring have a mean score greater than 29.5.

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Ionic bonding is called when you lose or gain electrons.

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