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3 years ago

341÷5 what's the remander

Mathematics

2 answers:

Maksim231197 [3]3 years ago

6 0

Answer:

its 1

Step-by-step explanation: 341/5= 68 r1

weqwewe [10]3 years ago

4 0

Answer:

68.2

Step-by-step explanation:

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Enter a fraction equavalent to 0.6. Use only whole numbers for numerators and denominators

hjlf

A fraction equivalent to 0.6 is
6/10= 3/5

6 0

2 years ago

(12+34)×2.3= (talk to me please please please)

serg [7]

Answer:

90.2

Step-by-step explanation:

thx for the follow!!!

5 0

2 years ago

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Suppose the expected tensile strength of type-A steel is 103 ksi and the standard deviation of tensile strength is 7 ksi. For ty

ExtremeBDS [4]

Answer:

i So the approximate distribution of $\= X$ is $\mu_{\= X} =103$ and $\sigma_{\= X} = 0.783$

ii So the approximate distribution of $\= Y$ is $\mu_{\= Y} =105$ and $\sigma_{\= Y} = 0.645$

the approximate distribution of $\=X - \= Y$ is $E (\= X - \= Y) = -2$ and $\sigma_{\= X - \=Y}=1.029$

Here we can see that the mean of the approximate distribution is negative which tell us that this negative value of the data for $\=X - \= Y$ sample are more and their frequency occurrence is higher than the positive values

the value of $P(-1 \le \=X - \= Y \le 1)$ is $= -0.1639$

Step-by-step explanation:

From the question we are given that

The expected tensile strength of the type A steel is $\mu_A = 103 ksi$

The standard deviation of type A steel is $\sigma_A = 7ksi$

The expected tensile strength of the type B steel is $\mu_B = 105\ ksi$

The standard deviation of type B steel is $\sigma_B = 5 \ ksi$

Also the assumptions are

Let $\= X$ be the sample average tensile strength of a random sample of 80 type-A specimens

Here $n_a =80$

Let $\= Y$ be the sample average tensile strength of a random sample of 60 type-B specimens.

Here $n_b = 60$

Let the sampling distribution of the mean be

$\mu _ {\= X} = \mu$

$=103$

Let the sampling distribution of the standard deviation be

$\sigma _{\= X} = \frac{\sigma }{\sqrt{n_a} }$

$= \frac{7}{\sqrt{80} }$

$=0.783$

So What this mean is that the approximate distribution of $\= X$ is $\mu_{\= X} =103$ and $\sigma_{\= X} = 0.783$

For $\= Y$

The sampling distribution of the sample mean is

$\mu_{\= Y} = \mu$

$= 105$

The sampling distribution of the standard deviation is

$\sigma _{\= Y} = \frac{\sigma }{\sqrt{n_b} }$

$= \frac{5}{\sqrt{60} }$

$= 0.645$

So What this mean is that the approximate distribution of $\= Y$ is $\mu_{\= Y} =105$ and $\sigma_{\= Y} = 0.645$

Now to obtain the approximate distribution for $\=X - \= Y$

$E (\= X - \= Y) = E (\= X) - E(\= Y)$

$= \mu_{\= X} - \mu_{\= Y}$

$= 103 -105$

$= -2$

The standard deviation of $\=X - \= Y$ is

$\sigma_{\= X - \=Y} = \sqrt{\sigma_{\= X}^2 - \sigma_{\= Y}^2}$

$= \sqrt{(0.783)^2 + (0.645)^2}$

$=1.029$

Now to find the value of $P(-1 \le \=X - \= Y \le 1)$

Let us assume that $F = \= X - \= Y$

$P(-1 \le F \le 1)$ $= P [\frac{-1 -E (F)}{\sigma_F} \le Z \le \frac{1-E(F)}{\sigma_F} ]$

$= P[\frac{-1-(-2)}{1.029} \le Z \le \frac{1-(-2)}{1.029} ]$

$= P[0.972 \le Z \le 2.95]$

$= P(Z \le 0.972) - P(Z \le 2.95)$

Using the z-table to obtain their z-score

$= 0.8345 - 0.9984$

$= -0.1639$

3 0

2 years ago

PLEASE HELP ME!

Firdavs [7]

Answer:

Step-by-step explanation:

educational guess ;)

8 0

3 years ago

Read 2 more answers

2) This is a line segment between the center and a point on the circle or sphere.

Dima020 [189]

Answer:

Radius

Step-by-step explanation:

A circle is made of specific components which define its shape. It has a fixed center which is equidistant apart from all points at its edge. The distance from the center to each point is called its radius. The distance from its edge through its center to the opposite edge is called the diameter.

3 0

2 years ago

341÷5 what's the remander​

341÷5 what's the remander