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2 years ago

Jenny has 36 cupcakes. How many kids (including Jenny) can share these cupcakes evenly?

Mathematics

2 answers:

ratelena [41]2 years ago

8 0

36 is the correct answer

irina1246 [14]2 years ago

7 0

36

36/?

Smallest amount of kids per cupcake is 1

36/1

36 kids

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Solve the inequality 15t > 180

bagirrra123 [75]

Answer:

t>12

Step-by-step explanation:

t>180/15

then divide 180 by 15

and you gey t>12

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3 years ago

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wolverine [178]

Answer:

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Step-by-step explanation:

f(x) -> 2 as x -> -∞ and f(x) -> ∞ as x -> ∞

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2 years ago

Simplify the expression completely. 16x-12x

romanna [79]

These are like terms, meaning that they are completely the same disregarding the coefficient.

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3 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

I am Lyosha [343]

Answer:

(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

Step-by-step explanation:

Let X = number of students who read above the eighth grade level.

(a)

A sample of n = 269 students are selected. Of these 269 students, X = 224 students who can read above the eighth grade level.

Compute the proportion of students who can read above the eighth grade level as follows:

$\hat p=\frac{X}{n}=\frac{224}{269}=0.8327$

The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

$1-\hat p=1-0.8327$

$=0.1673$

Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

n = 709

X = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

$\hat q=1-\hat p$

$=1-\frac{X}{n}$

$=1-\frac{546}{709}$

$=0.2299\\\approx 0.229$

The critical value of z for 95% confidence interval is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the 95% confidence interval for the population proportion as follows:

$CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$=0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)$

Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

5 0

3 years ago

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Allushta [10]

Answer:

108

Step-by-step explanation:

6 0

3 years ago

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