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3 years ago

If a cow has a mass of 9 × 10^2

Mathematics

2 answers:

kotykmax [81]3 years ago

5 0

Answer:

a) The blue whale has about 200 times more mass

Step-by-step explanation:

(blue whale mass)/(cow mass) = (1.8×10^5 kg)/(9×10^2 kg)

... = (1.8/9)×10^(5-2)

... = 0.2×10^3

... = 200

The blue whale has about 200 times as much mass as a cow.

FrozenT [24]3 years ago

4 0

Answer:

a) The blue whale has about 200 times more mass.

Step-by-step explanation:

A cow has a mass of 9 × 10^2 kilograms, and a blue whale has a mass of 1.8 × 10^5 kilograms.

The blue whale is about $\frac{1.8\times10^{5}}{9\times10^{2}}$ times more than cow.

= $\frac{180000}{900}$

= 200

Hence, the answer is a) The blue whale has about 200 times more mass.

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If f(×)=1/3×^2+5find f(-9)

Snezhnost [94]

Answer:

I think this is the answer

$f( - 9) = \frac{1}{3( - 9) {}^{2} } + 5 \\ = \frac{1216}{243} \\ = 5.0041$

5 0

3 years ago

Select all the values that cannot be probabilities a. 3/5b. 1c. -0.45d. 0.06e. 1.16f. 0g. 5/3h. √2

nikitadnepr [17]

Answer:

c,-0.45 ,

e. 1.16

g. 5/3

h. √2

Cannot be probability.

Step-by-step explanation:

Probability of any occurrence is always in the interval of 0 to 1. 0 and 1 inclusive.

0≤P(A)≤1

Any value outside this boundaries cannot be probability.

Therefore, -0.45,1.16,5/3 and √2 cannot be probability.

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3 years ago

What is 60 divided by 1.79

Marizza181 [45]

--->33.5195530726257

3 0

3 years ago

A random sample of 64 observations produced a mean value of 84 and standard deviation of 5.5. The 90% confidence interval for th

ASHA 777 [7]

Answer: The 90% confidence interval for the population mean μ is between 82.85 and 85.15,

Step-by-step explanation:

When population standard deviation is not given ,The confidence interval population proportion is given by ( $\mu$ ):-

$\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}$

, where n= Sample size.

s= Sample standard deviation

$\overline{x}$ = sample mean

t* = Critical t-value (Two-tailed)

As per given , we have

$\overline{x}=84$

n= 64

Degree of freedom : df = n-1=63

s= 5.5

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

Two-tailed T-value for df = 63 and $\alpha=1-0.90=0.1$ would be

$t_{\alpha/2,df}=t_{0.05,63}=1.669$ (By t-distribution table)

i.e. t*= 1.669

The 90% confidence interval for the population mean μ would be

$84\pm (1.669)\dfrac{5.5}{\sqrt{64}}$

$=84\pm (1.669)\dfrac{5.5}{8}$

$\approx84\pm 1.15$

$=(84-1.15,\ 84+1.15)=(82.85,\ 85.15)$

∴ The 90% confidence interval for the population mean μ is between 82.85 and 85.15,

6 0

3 years ago

Steve has three boxes. Box A has dimensions that are 5cm by 2cm by 3cm. Box B has dimensions that are 2cm by 2cm by 6cm. Box C h

Vedmedyk [2.9K]

Box A : 5 x 2 x 3 = 30 cubic cm

Box B: 2 x 2 x 6 = 24 cubic cm

Box C: 6 x 3 x 2 = 30 cubic cm

5 0

3 years ago