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2 years ago

What is 0.3,0.13,0.19 and 0.31 in order from least to greatest

Mathematics

2 answers:

Alexxx [7]2 years ago

6 0

0.3, 0.13,0.19,0.31 is the answer I'm pretty sure!!!!!!

wariber [46]2 years ago

3 0

0.3 then 0.13 then 0.19 then 0.31
have a nice day ;)

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Find the hupotenuse if the legs are 3 and 4

natita [175]

Answer:

None of these

Step-by-step explanation:

$hypotenuse \\ = \sqrt{ {3}^{2} + {4}^{2} } \\ = \sqrt{9 + 16} \\ = \sqrt{25} \\ = 5$

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3 years ago

Read 2 more answers

Find area and perimeter

alisha [4.7K]

Answer:

Perimeter

P = P1+P2

P1 = 4a P2=2(a+b)

P1=4*5 P2=2(3+5)

P1=20 P2=2*8

P2=16

P = 20+16

P=36

Area

A=A1+A2

A1= $a^{2}$ A2=a*b

A1= $5^{2}$ A2=3*5

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2 years ago

Suppose that a sample of size 100 is to be drawn from a population with standard deviation L0. a. What is the probability that t

NARA [144]

Answer:

The probability that the sample mean will lie within 2 values of μ is 0.9544.

Step-by-step explanation:

Here

the sample size is given as 100
the standard deviation is 10

The probability that the sample mean lies with 2 of the value of μ is given as

$P(| \bar{X}-\mu|$

Here converting the values in z form gives

$P(-2$

Substituting values

$P(-2$

From z table

$P(z\leq 2)=0.9772\\P(z\leq -2)=0.0228\\P(-2\leq z\leq 2)=P(z\leq 2)-P(z\leq -2)\\P(-2\leq z\leq 2)=0.9772-0.0228\\P(-2\leq z\leq 2)=0.9544\\$

So the probability that the sample mean will lie within 2 values of μ is 0.9544.

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2 years ago

The value of the 4th term is 80. the sequence is being doubled each step what is the 7th term?

leva [86]

This is a geometric sequence with common ratio (r) = 2.
an = ar^(n - 1)
a4 = a(2)^(4 - 1) = a(2)^3 = 8a
8a = 80
a = 80/8 = 10

7th term (a7) = 10(2)^(7 - 1) = 10(2^6) = 10(64) = 640

8 0

2 years ago

The weight of the chocolate and Hershey Kisses are normally distributed with a mean of 4.5338 G and a standard deviation of 0.10

Salsk061 [2.6K]

For the bell-shaped graph of the normal distribution of weights of Hershey kisses, the area under the curve is 1, the value of the median and mode both is 4.5338 G and the value of variance is 0.0108.

In the given question,

The weight of the chocolate and Hershey Kisses are normally distributed with a mean of 4.5338 G and a standard deviation of 0.1039 G.

We have to find the answer of many question we solve the question one by one.

From the question;

Mean(μ) = 4.5338 G

Standard Deviation(σ) = 0.1039 G

(a) We have to find for the bell-shaped graph of the normal distribution of weights of Hershey kisses what is the area under the curve.

As we know that when the mean is 0 and a standard deviation is 1 then it is known as normal distribution.

So area under the bell shaped curve will be

$\int\limits^{\infty}_{-\infty} {f(x)} \, dx$ = 1

This shows that that the total area of under the curve.

(b) We have to find the median.

In the normal distribution mean, median both are same. So the value of median equal to the value of mean.

As we know that the value of mean is 4.5338 G.

So the value of median is also 4.5338 G.

In the normal distribution mean, mode both are same. So the value of mode equal to the value of mean.

As we know that the value of mean is 4.5338 G.

So the value of mode is also 4.5338 G.

(d) we have to find the value of variance.

The value of variance is equal to the square of standard deviation.

So Variance = $(0.1039)^2$

Variance = 0.0108

Hence, the value of variance is 0.0108.

To learn more about normally distribution link is here

brainly.com/question/15103234

#SPJ1

3 0

6 months ago