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dmitriy555 [2]
2 years ago
7

han made some hot choclate by mixing 4 cups of milk with 6 tablespoons of cocoa. how many cups of milk per tablespoon of cocoa i

s that
Mathematics
2 answers:
QveST [7]2 years ago
7 0
I believe the answer is .68 cups because, 4 divided by 6 is about .68. You are trying to find the unit rate.
gavmur [86]2 years ago
4 0
The answer would be 1.5 cups of milk
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What is 50 divided by 48 in all kinds of forms
galben [10]
You put it 48 divide 50 which equally 0.96 or 50 divide 48 which equal 1.04166666667

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2 years ago
Kayden is buying bagels for a family gathering. Each bagel costs $1.50. Answer the questions below regarding the relationship be
Anna11 [10]

Answer:

y= $1.50x

Step-by-step explanation:

Each bagel is $1.50,therefore for each family member(x) the amount increaes by the same rate

4 0
3 years ago
Are the ratios 14:18 and 7:9 equivalent
Pepsi [2]

Answer:

Yes

Step-by-step explanation:

14/18 equals 7/9

Dividing by two on the first ratio makes:

7/9 equals 7/9

4 0
3 years ago
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Use base ten blocks to solve 53/4
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6 0
3 years ago
In a program designed to help patients stop smoking. 192 patients were given to sustained care, and 80.2% of them were no longer
stira [4]

Answer:

We conclude that 80% of patients stop smoking when given sustained care.

Step-by-step explanation:

We are given that in a program designed to help patients stop smoking. 192 patients were given to sustained care, and 80.2% of them were no longer smoking after one month.

Let p = <u><em>percentage of patients stop smoking when given sustained care.</em></u>

So, Null Hypothesis, H_0 : p = 80%     {means that 80% of patients stop smoking when given sustained care}

Alternate Hypothesis, H_A : p \neq 80%     {means that different from 80% of patients stop smoking when given sustained care}

The test statistics that would be used here <u>One-sample z test for proportions</u>;

                        T.S. =  \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of patients who stop smoking when given sustained care = 80.2%

           n = sample of patients = 192

So, <u><em>the test statistics</em></u>  =  \frac{0.802-0.80}{\sqrt{\frac{0.802(1-0.802)}{192} } }

                                     =  0.07

The value of z test statistics is 0.07.

<u></u>

<u>Also, P-value of the test statistics is given by the following formula;</u>

                P-value = P(Z > 0.07) = 1 - P(Z \leq 0.07)

                              = 1 - 0.52790 = 0.4721

<u>Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.</u>

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that 80% of patients stop smoking when given sustained care.

8 0
3 years ago
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