All categories

3 years ago

I need help with 3&4 plz

Mathematics

1 answer:

ira [324]3 years ago

6 0

Answer:

-5 and 8

in question 3and 4 we have to evaluate

You might be interested in

Please help asap! finding mean and range

stepan [7]

Mean = 26.75
Range = 25
The answer is C!

5 0

3 years ago

Read 2 more answers

Triangle X Y Z is shown. The length of side X Y is (9 n + 12) feet and the length of side X Z is (15 n minus 6) feet. Angles X Y

zmey [24]

Answer:

$n = 3$

$XY = 39$

$XZ = 39$

Step-by-step explanation:

Given

$XY = 9n + 12$

$XZ = 15n - 6$

Solving (a): n

To solve for n;

We make use of

$XY = XZ$

Substitute values for XY and XZ

$9n + 12 = 15n - 6$

$9n - 15n = -12 - 6$

$-6n = -18$

Divide through by -6

$n = 3$

Solving (b) XY

$XY = 9n + 12$

Substitute 3 for n

$XY = 9 * 3 + 12$

$XY = 27 + 12$

$XY = 39$

Solving (c): XZ

$XZ = XY$

$XZ = 39$

7 0

3 years ago

Find the quotient below. Write the answer as a fraction in simplest form. 1/3÷4 3/4Find the quotient below. Write the answer as

evablogger [386]

Answer:

Step-by-step explanation:

8 0

3 years ago

How can we put 45 rabbits in 9 cells, so that there are different number of rabbits in each cell and none of the cells is empty?

erica [24]

You can put one rabbit in cell one. Two rabbits in cell two. Three rabbits in cell three. Four rabbits in cell four. Five rabbits in cell five. Six rabbits in cell six. Seven rabbits in cell seven. Eight rabbits in cell eight. And finally nine rabbits in cell nine.

3 0

4 years ago

Read 2 more answers

A rare form of malignant tumor occurs in 11 children in a million, so its probability is 0.000011. Four cases of this tumor occ

Shkiper50 [21]

Answer:

a) The mean number of cases is 0.14608 cases.

b) The probability that the number of cases is exactly 0 or 1 is 0.990.

c) The probability of more than one case is 0.010

d) No, because the probability of more than one case is very small

Step-by-step explanation:

We can model this problem with a Poisson distribution, with parameter:

$\lambda=r*t=0.000011*13,280=0.14608$

a) The mean amount of cases is equal to the parameter λ=0.14608.

b) The probability of having 0 or 1 cases is:

$P(k=0)=\frac{\lambda^0 e^{-\lambda}}{0!}=\frac{1*0.864}{1} =0.864\\\\ P(k=1)=\frac{\lambda^1 e^{-\lambda}}{0!}=\frac{0.14608*0.864}{1} =0.126\\\\P(k\leq1)=0.864+0.126=0.990$

c) The probability of more than one case is:

$P(k>1)=1-P(k\leq 1)=1-0.990=0.010$

d) The cluster of 4 cases can not be due to pure chance, as it is a very high proportion of cases according to the average rate. Just having more than one case has a probability of 1%.

7 0

3 years ago