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Nina [5.8K]
2 years ago
8

What is the meaning of Post war?

Mathematics
1 answer:
Licemer1 [7]2 years ago
3 0
Post war means the time after a war had just taken place
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​It takes an ant farm 6 days to consume 3/8 of an apple. ​(​Show all work) At that rate, in how many days will the ant farm cons
Lana71 [14]

Answer: 48 days

Step-by-step explanation:

Given

Ant takes 6 days to consume  \dfrac{3}{8} of an apple

Using unitary method

6\ \text{days}\rightarrow \quad \frac{3}{8}\ \text{of an apple}\\\\1\ \text{apple takes}\rightarrow \quad \frac{8}{3}\times 6=16\ \text{days}\\\\3\ \text{apples take}\rightarrow\quad 16\times3=48\ \text{days}

6 0
3 years ago
13. which inequality has the following graphed solution?
mafiozo [28]
B. You have to solve for X, and graph using whatever is less than or greater than. 4x + 12 < 4 to 4x < -8 to x < -2.
5 0
2 years ago
The heights in inches of orangutans Are normally distributed with a population standard deviation of 3" and an unknown populatio
CaHeK987 [17]

Answer: The 95% confidence interval is approximately (55.57, 58.43)

======================================================

Explanation:

At 95% confidence, the z critical value is about z = 1.960 which you find using a table or a calculator.

The sample size is n = 17

The sample mean is xbar = 57

The population standard deviation is sigma = 3

The lower bound of the confidence interval is

L = xbar - z*sigma/sqrt(n)

L = 57 - 1.960*3/sqrt(17)

L = 55.5738905247863

L = 55.57

The upper bound is

U = xbar + z*sigma/sqrt(n)

U = 57 + 1.960*3/sqrt(17)

U = 58.4261094752137

U = 58.43

Therefore the confidence interval (L, U) turns into (55.57, 58.43) which is approximate.

7 0
3 years ago
Independent random samples of vehicles traveling past a given point on an interstate highway have been observed on monday versus
quester [9]
Hi! 

To compare this two sets of data, you need to use a t-student test:

You have the following data:

-Monday n1=16; <span>x̄1=59,4 mph; s1=3,7 mph

-Wednesday n2=20;  </span>x̄2=56,3 mph; s2=4,4 mph

You need to calculate the statistical t, and compare it with the value from tables. If the value you obtained is bigger than the tabulated one, there is a statistically significant difference between the two samples.

t= \frac{X1-X2}{ \sqrt{ \frac{(n1-1)* s1^{2}+(n2-1)* s2^{2} }{n1+n2-2}} * \sqrt{ \frac{1}{n1}+ \frac{1}{n2}} } =2,2510

To calculate the degrees of freedom you need to use the following equation:

df= \frac{ (\frac{ s1^{2}}{n1} + \frac{ s2^{2}}{n2})^{2}}{ \frac{(s1^{2}/n1)^{2}}{n1-1}+ \frac{(s2^{2}/n2)^{2}}{n2-1}}=33,89≈34

The tabulated value at 0,05 level (using two-tails, as the distribution is normal) is 2,03. https://www.danielsoper.com/statcalc/calculator.aspx?id=10

So, as the calculated value is higher than the critical tabulated one, we can conclude that the average speed for all vehicles was higher on Monday than on Wednesday.



5 0
3 years ago
Solve the system of equations.<br><br><br> 2x + 3y = 1<br> y = 3x + 15
MrMuchimi
Since y=3x+15
sub 3x+15 for y in other equation

2x+3(3x+15)=1
2x+9x+45=1
minus 45
11x=-44
divide 11
x=-4

sub
y=3x+15
y=3(-4)+15
y=-12+15
y=3

(-4,3)
6 0
2 years ago
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