All categories

3 years ago

Which measurement is most accurate to describe the weight of a baby

Mathematics

2 answers:

artcher [175]3 years ago

5 0

Answer:

Option 4). 4.5 kg

Step-by-step explanation:

The given options for the weight of a baby are

1) 5000 mg = 5000/1000 gms = 5 gms = .005 kg

2) 1400 g = 1.5 kg

3) 350 mg = 0.35 gms = $0.35(10^{-3}) kg$

4) 4.5 kg

Out of these options option 4 is the most suitable weight that describes the weight of a baby.

Lapatulllka [165]3 years ago

5 0

<h2>Answer:</h2>

<u>The right choice is </u><u>(D) 4.5 kg</u>

<h2>Step-by-step explanation:</h2>

If we look at the choices given above

5000 mg = 0.005kg which is not possible for the weight of a baby.

Similarly 1400 kg equals 1.4 kg which also doesn't make sense. In the same way 350 mg is also very less value for the weight of a baby so the only option that is suitable is 4.5kg which is option D

You might be interested in

A school purchsed basketball for $20 each and footballs for $15 each. they a brought a total of 28 for $480 how many of each typ

satela [25.4K]

Answer:

i in total spent : 480

Step-by-step explanation:

he bought 12 basketballs = 240

he bought 16 football = 240

you can decide how to make your system

4 0

3 years ago

I stink at math please help me!

IceJOKER [234]

Only 0.15 or 3/20 of voters were eligible for Shelly. Only 0.3125 or 5/16 of voters were eligible for Morgan.

6 0

3 years ago

What is 7 billion in standard form?

zimovet [89]

If by standard from you mean numbers (123) and not words then

1 milllion=1000 thousand or 1000*1000 or 1,000,000
1 billion=1000 million=1000*1000*1000=1,000,000,000
so 7billion=7*1000*1000*1000=7,000,000,000

4 0

3 years ago

Read 2 more answers

A small lawnmower company produced 1,500 lawnmowers in 2008. In an effort to determine how maintenance-free these units were, th

ikadub [295]

Answer:

The 95% confidence interval for the average number of years until the first major repair is (3.1, 3.5).

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the average using the finite correction factor is:

$CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\cdot\sqrt{\frac{N-n}{N-1}}$

The information provided is:

$N=1500\\n=183\\\sigma=1.47\\\bar x=3.3$

The critical value of <em>z</em> for 95% confidence level is,

<em>z</em> = 1.96

Compute the 95% confidence interval for the average number of years until the first major repair as follows:

$CI=\bar x\pm z_{\alpha/2}\cdot\frac{\sigma}{\sqrt{n}}\cdot\sqrt{\frac{N-n}{N-1}}$

$=3.3\pm 1.96\times\frac{1.47}{\sqrt{183}}\times\sqrt{\frac{1500-183}{1500-1}}\\\\=3.3\pm 0.19964\\\\=(3.10036, 3.49964)\\\\\approx (3.1, 3.5)$

Thus, the 95% confidence interval for the average number of years until the first major repair is (3.1, 3.5).

7 0

3 years ago

Lotteries are an important income source for various governments around the world. However, the availability of lotteries and ot

SashulF [63]

Answer: 0.0473

Step-by-step explanation:

Given : The proportion of gambling addicts : p=0.30

Let x be the binomial variable that represents the number of persons are gambling addicts.

with parameter p=0.30 , n= 10

Using Binomial formula ,

$P(x)=^nC_xp^x(1-p)^{n-x}$

The required probability = $P(x>5)=P(6)+P(7)+P(8)+P(9)+P(10)\\\\=^{10}C_6(0.3)^6(0.7)^{4}+^{10}C_7(0.3)^7(0.7)^{3}+^{10}C_8(0.3)^8(0.7)^{2}+^{10}C_9(0.3)^9(0.7)^{1}+^{10}C_{10}(0.3)^{10}(0.7)^{0}\\\\=\dfrac{10!}{4!6!}(0.3)^6(0.7)^{4}+\dfrac{10!}{3!7!}(0.3)^7(0.7)^{3}+\dfrac{10!}{2!8!}(0.3)^8(0.7)^{2}+\dfrac{10!}{9!1!}(0.3)^9(0.7)^{1}+(1)(0.3)^{10}\\\\=0.0473489874\approx0.0473$

Hence, the required probability = 0.0473

5 0

3 years ago