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larisa [96]
2 years ago
14

14 birds were sitting in a tree 21 more birds flew up to the tree how many birds were there altogether in the tree ?

Mathematics
2 answers:
kipiarov [429]2 years ago
8 0
They are now 35 birds sitting in a tree.
Bezzdna [24]2 years ago
3 0
There are 35 birds as 21 plus 12 is 35 so 35 birds are sitting on the tree
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A cyclist travels 4 miles in 20 minutes. At this rate, how many miles does the cyclist travel in 30 minutes?​
marshall27 [118]

Answer:

6

Step-by-step explanation:

20 mins. divided by 4 is 1mi./5 minute

so 1 miles in 5 minute

30 minutes divided by 5 is 6

so 6 miles in 6 minutes

6 0
2 years ago
Given ED, DB which statements about the figure are true? Check all that apply.
Klio2033 [76]

Answer: EB is bisected by DF

A is the midpoint DF

EB is a segment bisector

FA=1/2FC.

Step-by-step explanation:

6 0
1 year ago
Which function is shown in the graph below?
Stolb23 [73]
The annswer is y=log3x
3 0
1 year ago
Read 2 more answers
A new drug on the market is known to cure 30% of patients with cervical cancer. If a group of 18 patients is randomly selected,
Dima020 [189]

Answer:

Probability cured of cervical cancer =  18C0 (0.30)⁰(0.70)¹⁸ +  18C1(0.30)(0.70)¹⁷

Step-by-step explanation:

Given:

Patients cured = 30% = 0.30

Number of patients (n) = 18

Probability cured of cervical cancer = P(X≤1)

Probability cured of cervical cancer  = P(X=0) + P(X=1)

Probability cured of cervical cancer =  18C0 (0.30)⁰(0.70)¹⁸ +  18C1(0.30)(0.70)¹⁷

5 0
1 year ago
Suppose the area that can be painted using a single can of spray paint is slightly variable and follows a nearly normal distribu
Svetllana [295]

Answer:

Step-by-step explanation:

Hello!

You have the variable

X: Area that can be painted with a can of spray paint (feet²)

The variable has a normal distribution with mean μ= 25 feet² and standard deviation δ= 3 feet²

since the variable has a normal distribution, you have to convert it to standard normal distribution to be able to use the tabulated accumulated probabilities.

a.

P(X>27)

First step is to standardize the value of X using Z= (X-μ)/ δ ~N(0;1)

P(Z>(27-25)/3)

P(Z>0.67)

Now that you have the corresponding Z value you can look for it in the table, but since tha table has probabilities of P(Z, you have to do the following conervertion:

P(Z>0.67)= 1 - P(Z≤0.67)= 1 - 0.74857= 0.25143

b.

There was a sample of 20 cans taken and you need to calculate the probability of painting on average an area of 540 feet².

The sample mean has the same distribution as the variable it is ariginated from, but it's variability is affected by the sample size, so it has a normal distribution with parameners:

X[bar]~N(μ;δ²/n)

So the Z you have to use to standardize the value of the sample mean is Z=(X[bar]-μ)/(δ/√n)~N(0;1)

To paint 540 feet² using 20 cans you have to paint around 540/20= 27 feet² per can.

c.

P(X≤27) = P(Z≤(27-25)/(3/√20))= P(Z≤2.98)= 0.999

d.

No. If the distribution is skewed and not normal, you cannot use the normal distribution to calculate the probabilities. You could use the central limit theorem to approximate the sampling distribution to normal if the sample size was 30 or grater but this is not the case.

I hope it helps!

4 0
2 years ago
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