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Karolina [17]
3 years ago
11

LOGIC In an asteroid field, 75% of the asteroids are carbonaceous asteroids. There are 375,000 carbonaceous asteroids in the ast

eroid field. How many asteroids are not carbonaceous?
_________ asteroids are not carbonaceous.
Mathematics
2 answers:
Vlad1618 [11]3 years ago
5 0

Answer:

93,750

Step-by-step explanation:

icang [17]3 years ago
3 0

Answer:

?

Step-by-step explanation:

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Renna pushed the button for the elevator to go up, but it would not move. The weight limit for the elevator is 450 kilograms, bu
navik [9.2K]
4 people getting off would only be 470 so It would be 5 people to get off.
4 0
3 years ago
A study is performed in San Antonio to determine whether the average weekly grocery bill per five-person family in the town is s
bogdanovich [222]

Answer:

No, the sample evidence is not statistically significant (P-value = 0.125).

To reject the null hypothesis, the P-value has to be smaller than the significance level, so the significance level to reject the null hypothesis has to be 0.125 or higher.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the average weekly grocery bill per five-person family in San Antonio is significantly different from the national average.

Then, the null and alternative hypothesis are:

H_0: \mu=131\\\\H_a:\mu\neq 131

The significance level is 0.05.

The sample has a size n=50.

The sample mean is M=133.474.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.193.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11.193}{\sqrt{50}}=1.583

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{133.474-131}{1.583}=\dfrac{2.474}{1.583}=1.563

The degrees of freedom for this sample size are:

df=n-1=50-1=49

This test is a two-tailed test, with 49 degrees of freedom and t=1.563, so the P-value for this test is calculated as (using a t-table):

P-value=2\cdot P(t>1.563)=0.125

As the P-value (0.125) is bigger than the significance level (0.05), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the average weekly grocery bill per five-person family in San Antonio is significantly different from the national average.

8 0
2 years ago
1: Create a single variable linear equation that has no solution. Solve the equation algebraically to prove that it does not hav
OLga [1]

Answer:

(1) y=C, C\neq0

(2) x=C

(1) y=0

Step-by-step explanation:

For the linear equation, the power on the solitary variable must be 1.

When the graph of a function crosses/ touches the x-axis than the equation has solution/solutions otherwise it does not has a solution.

When it touches/cresses the x-axis for infinite times then it will have an infinite number of solutions.

(1) The linear equation y-C=0, where y is a variable and C is a non-zero constant,  C \neq 0, has the graph parallel to the x-axis. It has no solution as the graph will never cross/touch the x-axis.

Algebraically, in linear equation y-C=0, is independent of x, so there is no value of x for the solution to be exist.

(2) The linear equation x-C=0, where x is a variable and C is a constant, has one solution.

The solution for the equation is

x-C=0

\Rightarrow x=C, where C\neq0.

(3) The linear equation y=0, where y is a variable, has the graph coinciding with the x-axis. So, it has infinitely many solutions.

Algebraically, in linear equation y=0, is independent of x, so for all value of x, the given equation is zero. Hence, there are infinitely many solutions.

3 0
3 years ago
What would be the missing angle?
Korvikt [17]

x = 60°

180 - 120 = 60

6 0
3 years ago
I need help.. with this worksheet
ipn [44]
8. $298 4. 250 9. 28 6. 8.7 5. 465
3 0
3 years ago
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