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2 years ago

What are compatible number of 7233 divided by 84

1 answer:

4 0

To me, do the math using compatible number, means to estimate
7233 can round to 8000
and 84 can round to 80
now, this makes it easy, you can do it mentally, which is 100

In more detailed,
7233 can round to 7225
and 84 can round to 85
then, (not so obvious, but however) 85 times 85 equal 7225

so the compatible number of 7233 divided by 84, is 7225 and 85

I know it's a bit late, but hope it helps

You might be interested in

A genetic experiment involving peas yielded one sample of offspring consisting of 437437 green peas and 129129 yellow peas. Use

navik [9.2K]

Answer:

The null hypothesis: $\mathbf{H_o: p=0.27}$

The alternative hypothesis: $\mathbf{H_1: p \neq 0.27}$

Test statistics : z = −2.30

P-value: = 0.02144

Decision Rule: Since the p-value is lesser than the level of significance; then we reject the null hypothesis.

Conclusion: We accept the alternative hypothesis and conclude that under the same circumstances the proportion of offspring peas will be yellow is not equal to 0.27

Step-by-step explanation:

From the given information:

Let's state the null and the alternative hypothesis;

Since The claim is that 27% of the offspring peas will be yellow.

The null hypothesis state that the proportion of offspring peas will be yellow is equal to 0.27.

i.e

$\mathbf{H_o: p=0.27}$

The alternative hypothesis state that the proportion of offspring peas will be yellow is not equal to 0.27

$\mathbf{H_1: p \neq 0.27}$

<u>The test statistics:</u>

we are given 437 green peas and 129 yellow apples;

Hence;

$\hat p = \dfrac{x}{n}$

where ;

$\hat p$ = sample proportion

x = number of success

n = total number of the sample size

$\hat p = \dfrac{129}{437+129}$

$\hat p = \dfrac{129}{566}$

$\mathbf{\hat p = 0.2279}$

Now; the test statistics can be computed as :

$z = \dfrac { \hat p -p }{\sqrt {\dfrac{p(1-p)}{n} } }$

$z = \dfrac {0.2279 -0.27 }{\sqrt {\dfrac{0.27(1-0.27)}{566} } }$

$z = \dfrac {-0.043 }{\sqrt {\dfrac{0.27(0.73)}{566} } }$

$z = \dfrac {-0.043 }{\sqrt {\dfrac{0.1971}{566} } }$

$z = \dfrac {-0.043 }{\sqrt {3.48233216*10^{-4} } }$

$z = \dfrac {-0.043 }{0.01866} }$

z = −2.30

C. P-value

P-value = P(Z < z)

P-value = P(Z< -2.30)

By using the P-value method and the normal distribution as an approximation to the binomial distribution.

from the table of standard normal distribution

move left until the first column is reached. Note the value as –2.0

move upward until the top row is reached. Note the value as 0.30

find the probability value as 0.010724 by the intersection of the row and column values gives the area to the left of

z = -2.30

P- value = 2P(z ≤ -2.30)

P-value = 2 × 0.01072

P - value = 0.02144

Decision Rule: Since the p-value is lesser than the level of significance; then we reject the null hypothesis.

Conclusion: We accept the alternative hypothesis and conclude that under the same circumstances the proportion of offspring peas will be yellow is not equal to 0.27

8 0

2 years ago

Can someone please help me with this, im doing a test and im stuck on this question.

Mashutka [201]

Answer:

Step-by-step explanation:

first do 75/50 = 1.5

then multiply red by 1.5

13 landed on red so

13 x 1.5 = 19.5

but since you round up it is 20

6 0

3 years ago

Find the value of X, Z and Y.

FromTheMoon [43]

Answer:

x = 5; y = 138, z= 42

Step-by-step explanation:

4x- 17 = 3 [ opposite sides of a parallelogram are equal]

4x = 3+ 17 = 20; X= 20/4 = 5

138 + Z = 180( angle on a straight line)

Z is an alternate angles to the remaining angle off 138 forming the straight line.

z= 180-138 = 42 degrees

Y = 138 [ opposite angles of a regular quadilareral been equal]

6 0

3 years ago

The acceleration, in meters per second per second, of a race car is modeled by A(t)=t^3−15/2t^2+12t+10, where t is measured in s

oksian1 [2.3K]

Answer:

The maximum acceleration over that interval is $A(6) = 28$ .

Step-by-step explanation:

The acceleration of this car is modelled as a function of the variable $t$ .

Notice that the interval of interest $0 \le t \le 6$ is closed on both ends. In other words, this interval includes both endpoints: $t = 0$ and $t= 6$ . Over this interval, the value of $A(t)$ might be maximized when $t$ is at the following:

One of the two endpoints of this interval, where $t = 0$ or $t = 6$ .
A local maximum of $A(t)$ , where $A^\prime(t) = 0$ (first derivative of $A(t)\!$ is zero) and $A^{\prime\prime}(t)$ (second derivative of $\! A(t)$ is smaller than zero.)

Start by calculating the value of $A(t)$ at the two endpoints:

$A(0) = 10$ .
$A(6) = 28$ .

Apply the power rule to find the first and second derivatives of $A(t)$ :

$\begin{aligned} A^{\prime}(t) &= 3\, t^{2} - 15\, t + 12 \\ &= 3\, (t - 1) \, (t + 4)\end{aligned}$ .

$\displaystyle A^{\prime\prime}(t) = 6\, t - 15$ .

Notice that both $t = 1$ and $t = 4$ are first derivatives of $A^{\prime}(t)$ over the interval $0 \le t \le 6$ .

However, among these two zeros, only $t = 1\!$ ensures that the second derivative $A^{\prime\prime}(t)$ is smaller than zero (that is: $A^{\prime\prime}(1) < 0$ .) If the second derivative $A^{\prime\prime}(t)\!$ is non-negative, that zero of $A^{\prime}(t)$ would either be an inflection point (if $A^{\prime\prime}(t) = 0$ ) or a local minimum (if $A^{\prime\prime}(t) > 0$ .)

Therefore $\! t = 1$ would be the only local maximum over the interval $0 \le t \le 6\!$ .

Calculate the value of $A(t)$ at this local maximum:

$A(1) = 15.5$ .

Compare these three possible maximum values of $A(t)$ over the interval $0 \le t \le 6$ . Apparently, $t = 6$ would maximize the value of $A(t)\!$ . That is: $A(6) = 28$ gives the maximum value of $\! A(t)$ over the interval $0 \le t \le 6\!$ .

However, note that the maximum over this interval exists because $t = 6\!$ is indeed part of the $0 \le t \le 6$ interval. For example, the same $A(t)$ would have no maximum over the interval $0 \le t < 6$ (which does not include $t = 6$ .)