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3 years ago

A number close to its actual value is a(n)

Mathematics

1 answer:

OLEGan [10]3 years ago

6 0

The answer to that question is estimate

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Some please help and explain the steps. Tysmmmmm

-Dominant- [34]

Answer:

since angle B and angle C contains the same arc AD. set the angles equal

11x-3=8x+15

x=6

so the angle has a measure of 63 degrees

the central angle is the angle of the arc

and it is always 2 times the angle that we just that contains the arc but does touches the end of the circle.

so its 63 *2 =126 degrees.

5 0

3 years ago

Please help!

Musya8 [376]

Answer:

See the attachment please.

Step-by-step explanation:

Statistics!! All work is on the picture, plus the boxed answers.

5 0

3 years ago

A used car dealer has 30 cars and 10 of them are lemons (i.e.~ mechanically faulty used cars), and you don't know which is which

Triss [41]

Answer: Hence, the probability that he will get at least one lemon is 0.70.

Step-by-step explanation:

Since we have given that

Number of cars = 30

Number of lemon cars = 10

Number of other than lemon cars = 30-10 = 20

According to question, he bought 3 cars,

we need to find the probability that you will get at least one lemon.

So, P(X≤1)=1-P(X=0)=1-P(no lemon)

Here, P(no lemon ) is given by

$\dfrac{20}{30}\times \dfrac{20}{30}\times \dfrac{20}{30}=(\dfrac{20}{30})^3$

so, it becomes,

$P(X\geq 1)=1-(\dfrac{20}{30})^3=1-(0.67)^3=0.70$

Hence, the probability that he will get at least one lemon is 0.70.

5 0

3 years ago

Find the HCF of 750,630 and 18900

Sever21 [200]

The answer is:
Hcf=150

3 0

3 years ago

A shoe manufacturer was investigating the weights of men's soccer cleats. He felt that the weight of these cleats was less than

aleksklad [387]

Answer:

The conclusion is that the researcher was correct

Step-by-step explanation:

From the question we are told that

The sample size is $n = 13$

The sample mean is $\= x = 9.63$

The standard deviation is $s = 0.585$

The significance level is $\alpha = 0.05$

The Null Hypothesis is $H_o : \mu = 0$

The Alternative Hypothesis is $H_a = \mu < 10$

The test statistic is mathematically represented as

$t = \frac{\= x - \mu }{\frac{s}{\sqrt{n} } }$

Substituting values

$t = \frac{9.63 - 10 }{\frac{0.585}{\sqrt{13} } }$

$t = - 2.280$

Now the critical value for $\alpha$ is

$t_{\alpha } = 1.645$

This obtained from the critical value table

So comparing the critical value of alpha and the test value we see that the test value is less than the critical value so the Null Hypothesis is rejected

The conclusion is that the researcher was correct

6 0

3 years ago