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3 years ago

Ella already knew 5 appetizer recipes before starting culinary school, and she will learn 2 new

Mathematics

1 answer:

Natalija [7]3 years ago

5 0

A = 2w+5

example: on week three, she would have the recipes from the other weeks that you have to account for as well as the extra five

2(3)+5=11 so A would equal 11 by week three.

Use multiplication because it is essentially repeated addition.

2(from week 1) + 2(from week 2) + 2(from week 3) + 5(prior to school) =11

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Which answer choice shows the decimal 23.476 written in expanded form?

Sergio039 [100]

Answer: 23.476 = 20+3+0.4+0.07+0.006

Step-by-step explanation:

I think you were supposed to write down the choices but I will guess the answer is this one:

23.476 = 20+3+0.4+0.07+0.006

Hope I helped!

6 0

2 years ago

HELP ASAP!!!

saul85 [17]

D.(6,3)

because your in put is not the same as any of the other in puts

3 0

3 years ago

Find the number of ways to distribute six different toys to three different children such that each child gets at least one toy.

Lisa [10]

Answer:

Step-by-step explanation:

Given that there are six different toys and they are to be distributed to three different children.

The restraint here is each child gets atleast one toy.

Let us consider the situation as this.

Since each child has to get atleast one toy no of ways to distribute

any 3 toys to the three children each. This can be done by selecting 3 toys from 6 in 6C3 ways and distributing in 3! ways

So 3 toys to each one in 6x5x4 =120 ways

Now remaining 3 toys can be given to any child.

Hence remaining 3 toys can be distributed in 3x3x3 =27 ways

Total no of ways

= 120(27)

= 3240

8 0

3 years ago

A basketball team played 45 games. they won 60% of them. how many did they win?

SVETLANKA909090 [29]

The answer that I got was 27, hope it helps

5 0

3 years ago

Read 2 more answers

Consider the following hypothesis test. H0: μ ≥ 55 Ha: μ < 55 A sample of 36 is used. Identify the p-value and state your con

Ket [755]

Answer:

Step-by-step explanation:

Given that:

$H_o: \mu \ge 55 \ \\ \\ H_1 : \mu < 55$

(a) For x = 54 and s = 5.3

The test statistics can be computed as:

$Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }$

$Z = \dfrac{54-55}{\dfrac{5.3}{\sqrt{36}} }$

$Z = \dfrac{-1}{\dfrac{5.3}{6} }$

Z = -1.132

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(-1.132,35,1) = 0.1326

Decision: p-value is greater than significance level; do not reject $\mathbf{H_o}$

$Conclusion: \ There \ is \ insufficient \ evidence \ to \ conclude \ that \$ $\mu < 55$

For x = 53 and s = 4.6

The test statistics can be computed as:

$Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }$

$Z = \dfrac{53-55}{\dfrac{4.6}{\sqrt{36}} }$

$Z = \dfrac{-2}{\dfrac{4.6}{6} }$

Z = -2.6087

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(-2.6087,35,1) =0.0066

Decision: p-value is < significance level; we reject the null hypothesis.

$Conclusion: \ There \ is \ sufficient \ evidence \ to \ conclude \ that$ $\mu < 55$

For x = 56 and s = 5.0

The test statistics can be computed as:

$Z = \dfrac{\overline x - \mu}{\dfrac{s}{\sqrt{n}} }$

$Z = \dfrac{56-55}{\dfrac{5.0}{\sqrt{36}} }$

Z = 1.2

degree of freedom = n - 1

= 36 - 1

= 35

Using the Excel Formula:

P-Value = T.DIST(1.2,35,1) = 0.88009

Decision: p-value is greater than significance level; do not reject $\mathbf{H_o}$

$Conclusion: \ There \ is \ insufficient \ evidence \ to \ conclude \ that \$ $\mu < 55$

6 0

3 years ago