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Mathematics

2 answers:

nordsb [41]3 years ago

4 0

Answer:

If the frame for the object is taller than the line of text in which it appears, text might overlap the imported image or you might see increased space above the line. Consider selecting a different anchored object position, inserting a soft or hard line break, resizing the inline object, or specifying a different leading value for the surrounding lines.

Step-by-step explanation:

this is what i could think of hope this helps

algol [13]3 years ago

3 0

Answer:

what do u mean

Step-by-step explanation:

You might be interested in

You charge people $10 per hour to pick fruits and $8 per hour

Aleks04 [339]

Fruits :
Y=10(h)

Flowers;
Y=8(h)

Given: h=6

Fruits:
10(6) = $60

Flowers:
8(6)= $48

Back to the question, how much more from picking fruits?

$60 - $48 = $12 more picking fruits.

Hope this helps!

8 0

3 years ago

NEED HELP LIKE NOW 15 points if answered!!!!!!!<br> how do you work it out

goblinko [34]

Remark

You have to stare at this for a moment before you catch on. When you fold the arc into a circle, 8pi becomes the circumference of the base circle.

Formula

C = 2*pi*r

C = 8pi

Solution

8pi = 2pi * r Divide both sides by 2pi

8pi / 2 pi = r

4 = r Answer

4 0

3 years ago

Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we

Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N( $\mu = 7 lbs , \sigma^{2} = 0.875^{2} lbs$ )

The standard normal z distribution is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, X bar = sample mean weight

n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{7.5-7}{\frac{0.875}{\sqrt{4} } }$ ) = P(Z > 1.1428) = 0.1283 (using z% table)