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2 years ago

6

Which could be the name of a point?1: A2: m3: <--> AB4: AB

2 answers:

ivanzaharov [21]2 years ago

5 0

Answer:

A die is rolled then write sample space *s* and number of sample point n(S).

Tanzania [10]2 years ago

3 0

Answer:

3

Step-by-step explanation:

I do believe hope this helps

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Drag equations into order to show a way to solve for x.<br> HELP!

Sholpan [36]

Answer:

Step-by-step explanation:

8x - 16 = 40

8x = 56

x = 7

I hope I helped you.

7 0

3 years ago

Twin brothers, Billy and Bobby, can mow their grandparent's lawn together in 53 minutes. Billy could mow the lawn by himself

podryga [215]

5353 minutes = 89.22 hours
Maybe the figures should be 53 and 20??

Let's let Billy = A
The formula to use for work problems is
Time = (A*B) / (A+B)

53 = ( (A +20)* A) / (A +20 + A) )
53 = ( A^2 + 20A ) / (2A + 20)
A^2 + 20A -106A -1060
A^2 -86A -1060
A = 96.935
Bobby Mows in 116.935 minutes

3 0

3 years ago

Look at the picture<br><br>

Fantom [35]

Answer:

The x-intercept is 2.

Step-by-step explanation:

$2 \sqrt[3]{x - 10} + 4 = 0$

$2 \sqrt[3]{x - 10} = - 4$

$\sqrt[3]{x - 10 } = - 2$

$x - 10 = - 8$

$x = 2$

3 0

2 years ago

12x5 – 10x2 + x – 45

qaws [65]

(a)
$12x ^{5} - 10x ^{2} + x - 45$
(b)we have 4 terms in this expression.
(c)+12 is the leading coefficient in
$12x ^{5}$

(d) constant is -45

6 0

3 years ago

Using the simple random sample of weights of women from a data set, we obtain these sample statistics: nequals45 and x overbar

Tomtit [17]

Answer: (141.1, 156.48)

Step-by-step explanation:

Given sample statistics : $n=45$

$\overline{x}=148.79\text{ lb}$

$\sigma=31.37\text{ lb}$

a) We know that the best point estimate of the population mean is the sample mean.

Therefore, the best point estimate of the mean weight of all women = $\mu=148.79\text{ lb}$

b) The confidence interval for the population mean is given by :-

$\mu\ \pm E$ , where E is the margin of error.

Formula for Margin of error :-

$z_{\alpha/2}\times\dfrac{\sigma}{\sqrt{n}}$

Given : Significance level : $\alpha=1-0.90=0.1$

Critical value : $z_{\alpha/2}=z_{0.05}=\pm1.645$

Margin of error : $E=1.645\times\dfrac{31.37}{\sqrt{45}}\approx 7.69$

Now, the 90% confidence interval for the population mean will be :-

$148.79\ \pm\ 7.69 =(148.79-7.69\ ,\ 148.79+7.69)=(141.1,\ 156.48)$

Hence, the 90% confidence interval estimate of the mean weight of all women= (141.1, 156.48)

3 0

3 years ago