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murzikaleks [220]
3 years ago
10

on an average a bakery sells 44 chocolate cakes every 4 days at this rate about how many chocolate cakes can the bakery sell in

8 days
Mathematics
1 answer:
zubka84 [21]3 years ago
7 0
Okay. First things first, we can find the unit rate of chocolate cakes that the bakery sells. 44 chocolate cakes are sold in 4 days. 44/4 is 11. 11 chocolate cakes are sold per day. Now, we can multiply that total by 8 to get the total for 8 days. If you know basic multiplication, this one's simple. 11 * 8 is 88. There. The bakery can sell 88 chocolate cakes in 8 days.
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Zanzabum

Answer:  x=-303

Step-by-step explanation:

x+310=7

We move all terms to the left:

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5 0
3 years ago
Read 2 more answers
Please help me with it​
NARA [144]

Answer:

Pythagorean Theorem:

a^2 + b^2 = c^2

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7 0
3 years ago
What is the base of a triangle that has a height of 6 centimeters and an area of 18 centimeters? Use the formula h = StartFracti
AveGali [126]

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3 years ago
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natali 33 [55]

Answer:

We conclude that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has​ increased or remained same.

Step-by-step explanation:

We are given that in a previous​ poll, 40​% of adults with children under the age of 18 reported that their family ate dinner together seven nights a week.

Suppose​ that, in a more recent​ poll, 456 of 1194 adults with children under the age of 18 reported that their family ate dinner together seven nights a week.

Let p = <u><em>proportion of families with children under the age of 18 who eat dinner together seven nights a week.</em></u>

SO, Null Hypothesis, H_0 : p \geq 40%      {means that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has​ increased or remained same}

Alternate Hypothesis, H_A : p < 40%      {means that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has​ decreased}

The test statistics that would be used here <u>One-sample z test for</u> <u>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 families with children under the age of 18 who eat dinner together seven nights a week = \frac{456}{1194} = 0.38

           n = sample of adults = 1194

So, <u><em>the test statistics</em></u>  =  \frac{0.38-0.40}{\sqrt{\frac{0.38(1-0.38)}{1194} } }

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The value of z test statistics is -1.424.

<u>Now, at 0.01 significance level the z table gives critical value of -2.326 for left-tailed test.</u>

Since our test statistic is more than the critical value of z as -1.424 > -2.326, 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 the proportion of families with children under the age of 18 who eat dinner together seven nights a week has​ increased or remained same.

8 0
4 years ago
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4 years ago
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