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3 years ago

Estimate 307‾‾‾‾√ between two consecutive whole numbers. Enter the numbers in the boxes below.

Mathematics

1 answer:

creativ13 [48]3 years ago

8 0

Answer:

17, 18

Step-by-step explanation:

closest perfect square that is lower than 307 is 289

closest perfect square that is greater than 307 is 324

$\sqrt{289} < \sqrt{307} < \sqrt{324}$

$17$

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Best Buy is giving you a free t.v. with purchase of a new refrigerator. The t.v. is 30 inches tall and 48 inches wide. How big i

user100 [1]

Answer:

57 inches

Step-by-step explanation:

The height, width, and diagonal of the TV screen form a right triangle.

Use the pythagorean theorem, a² + b² = c², and solve for c, the hypotenuse/diagonal of the TV.

a² + b² = c²

30² + 48² = c²

900 + 2304 = c²

3204 = c²

56.6 = c

Round this to the nearest inch:

56.6

= 57

So, the diagonal of the TV screen is 57 inches

5 0

2 years ago

If 6 bags of chocolate chips are needed to make 86 cookies, how many cookies can 8 bags chocolate chips make?

gulaghasi [49]

Answer:

114.666666667

Step-by-step explanation:

Step 1:

6/86 = 8/x Equation

Step 2:

6x = 688 Multiply

Step 3:

x = 688 ÷ 6 Divide

Answer:

x = 114.666666667

Hope This Helps :)

4 0

3 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

vredina [299]

Answer:

c) P(270≤x≤280)=0.572

d) P(x=280)=0.091

Step-by-step explanation:

The population of bearings have a proportion p=0.90 of satisfactory thickness.

The shipments will be treated as random samples, of size n=500, taken out of the population of bearings.

As the sample size is big, we will model the amount of satisfactory bearings per shipment as a normally distributed variable (if the sample was small, a binomial distirbution would be more precise and appropiate).

The mean of this distribution will be:

$\mu_s=np=500*0.90=450$

The standard deviation will be:

$\sigma_s=\sqrt{np(1-p)}=\sqrt{500*0.90*0.10}=\sqrt{45}=6.7$

We can calculate the probability that a shipment is acceptable (at least 440 bearings meet the specification) calculating the z-score for X=440 and then the probability of this z-score:

$z=(x-\mu_s)/\sigma_s=(440-450)/6.7=-10/6.7=-1.49\\\\P(z>-1.49)=0.932$

Now, we have to create a new sampling distribution for the shipments. The size is n=300 and p=0.932.

The mean of this sampling distribution is:

$\mu=np=300*0.932=279.6$

The standard deviation will be:

$\sigma=\sqrt{np(1-p)}=\sqrt{300*0.932*0.068}=\sqrt{19}=4.36$

c) The probability that between 270 and 280 out of 300 shipments are acceptable can be calculated with the z-score and using the continuity factor, as this is modeled as a continuos variable:

$P(270\leq x\leq280)=P(269.5$