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2 years ago

13

You have $25.00 to spend at the county fair. Admission to the fair is $5. The cost of a carnival ride ticket is $2. If you spend

all of your money at the fair, how many carnival ride tickets can you buy?

1 answer:

Gre4nikov [31]2 years ago

6 0

Answer:

10

Step-by-step explanation:

25 - 5 = 20

20 / 2 = 10

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Area of the blue shaded region??? please show work

konstantin123 [22]

Answer:

21

Step-by-step explanation:

you first look at the circle

next you see the numbers in the circle

after that you multiply the numbers together

so, 3 x 7 = 21

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3 years ago

Describe why you can change a subtraction problem into an addition problem

Elena-2011 [213]

A subtraction problem can be turned in to an addition problem because of negative numbers. Take 5-2=3 for example. If you change 5-2 into an addition problem then you would but he addition sign and a negative sign on 2. This will then leave you with the problem looking like 5+ -2. If you add 5 and -2 you will still get three.

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3 years ago

Find the length of the third side to the nearest tenth.

Aliun [14]

Answer:

5.3

Step-by-step explanation:

You use the Pythagoras theorem which is a^2+b^2=c^2.

Since you're solving for a side length that's not the hypotenuse you will manipulate the equation to c^2-a^2=b^2. From here you just plug in numbers.

8^2 - 6^2 = b^2

64 - 36 = b^2

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b = 5.2915.... = 5.3

8 0

3 years ago

Assume that hybridization experiments are conducted with peas having the property that for offspring, there is a 0.75 probabili

ivolga24 [154]

Answer:

(a) The mean and the standard deviation for the numbers of peas with green pods in the groups of 36 is 27 and 2.6 respectively.

(b) The significantly low values are those which are less than or equal to 21.8. And on the other hand, the significantly higher values are those which are greater than or equal to 32.2.

(c) The result of 15 peas with green pods is a result that is significantly low value.

Step-by-step explanation:

We are given that hybridization experiments are conducted with peas having the property that for offspring, there is a 0.75 probability that a pea has green pods.

Assume that the offspring peas are randomly selected in groups of 36.

The above situation can be represented as a binomial distribution;

where, n = sample of offspring peas = 36

p = probability that a pea has green pods = 0.75

(a) The mean of the binomial distribution is given by the product of sample size (n) and the probability (p), that is;

Mean, $\mu$ = n $\times$ p

= 36 $\times$ 0.75 = 27 peas

So, the mean number of peas with green pods in the groups of 36 is 27.

Similarly, the standard deviation of the binomial distribution is given by the formula;

Standard deviation, $\sigma$ = $\sqrt{n \times p \times (1-p)}$

= $\sqrt{36 \times 0.75 \times (1-0.75)}$

= $\sqrt{6.75}$ = 2.6 peas

So, the standard deviation for the numbers of peas with green pods in the groups of 36 is 2.6.

(b) Now, the range rule of thumb states that the usual range of values lies within the 2 standard deviations of the mean, that means;

$\mu - 2 \sigma$ = 27 - (2 $\times$ 2.6)

= 27 - 5.2 = 21.8

$\mu + 2 \sigma$ = 27 + (2 $\times$ 2.6)

= 27 + 5.2 = 32.2

This means that the significantly low values are those which are less than or equal to 21.8.

And on the other hand, the significantly higher values are those which are greater than or equal to 32.2.

(c) The result of 15 peas with green pods is a result that is a significantly low value because the value of 15 is less than 21.8 which is represented as a significantly low value.

5 0

3 years ago

<img src="https://tex.z-dn.net/?f=%20%5Csqrt%5B4%5D%7B%20lnx%20%7D%20" id="TexFormula1" title=" \sqrt[4]{ lnx } " alt=" \sqrt[4]

Kamila [148]

Use the power rule for differentiation:

$\dfrac{\text{d}}{\text{d}x} (f(x))^k = k(f(x))^{k-1}f'(x)$

You can use this formula if you remember that a root is just a rational exponential:

$\sqrt[4]\ln(x) = (\ln(x))^{\frac{1}{4}}$

So, remembering that the derivative of the logarithm is 1/x, you have

$\dfrac{\text{d}}{\text{d}x} (\ln(x))^{\frac{1}{4}} = \frac{1}{4}(\ln(x))^{\frac{1}{4}-1}\dfrac{1}{x}$

Which you can rewrite as

$\dfrac{1}{4}(\ln(x))^{\frac{1}{4}-1}\dfrac{1}{x} =\dfrac{1}{4}(\ln(x))^{\frac{-3}{4}}\dfrac{1}{x} =\dfrac{1}{4}\dfrac{1}{\sqrt[4]{\ln(x))^3}}\dfrac{1}{x} = \dfrac{1}{4x\sqrt[4]{\ln(x))^3}}$

3 0

3 years ago