Teresa and Susan have been given with four cards with four different functions on them.

All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be

Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a student is proficient in reading but not mathematics and $A \cap B$ is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

$B = b + (A \cap B)$

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

$(A \cup B) + C = 1$

In which

$(A \cup B) = a + b + (A \cap B)$

65% were found to be proficient in both reading and mathematics.

This means that $A \cap B = 0.65$

78% were found to be proficient in mathematics

This means that $B = 0.78$

$B = b + (A \cap B)$

$0.78 = b + 0.65$

$b = 0.13$

85% of the students were found to be proficient in reading

This means that $A = 0.85$

$A = a + (A \cap B)$

$0.85 = a + 0.65$

$a = 0.20$

Proficient in at least one:

$(A \cup B) = a + b + (A \cap B) = 0.20 + 0.13 + 0.65 = 0.98$

What is the probability that the student is proficient in neither reading nor mathematics?

$(A \cup B) + C = 1$

$C = 1 - (A \cup B) = 1 - 0.98 = 0.02$

There is a 2% probability that the student is proficient in neither reading nor mathematics.

6 0

2 years ago

A random sample of 35 undergraduate students who completed two years of college were asked to take a basic mathematics test. The

disa [49]

Answer:

The 90 % confidence limits are (-2.09, 8.09).

Since the calculated values do not lie in the critical region we accept our null hypothesis.

Step-by-step explanation:

The null and alternative hypothesis are given by

H0: σ₁²= σ₂² against Ha: σ₁² ≠ σ₂²

Confidence interval for the population mean difference is given by

(x`1- x`2) ± t √S²(1/n1 + 1/n2)

Where S ²= (n1-1)S₁² + S²₂(n2-1)/n1+n2-2

Critical value of t with n1+n2-2= 50+ 35-2= 83 will be -1.633

Now calculating

S ²=34* (12.8)²+ (14.6)²*49/83= 192.96

Now putting the values in the t- test

(75.1 -72.1) ± 1.633 √ 192.96(1/35 +1/50)

=3 ± 5.09

=-2.09, 8.09 is the 90 % confidence interval for the difference

The 90 % confidence limits are (-2.09, 8.09).

Since the calculated values do not lie in the critical region we accept our null hypothesis.

5 0

2 years ago

A man walks at 60km/ hr. How long does he walks between 9:30 am and 12 noon

N76 [4]

Answer:

He walks 150 km in 2.5 hours.

Step-by-step explanation:

From 9:30 a.m. to noon, it's 2.5 hours.

distance = speed * time

distance = 60 km/h * 2.5 h

distance = 150 km

Answer: He walks 150 km in 2.5 hours.

3 0

2 years ago

Is 3x+4y a polynomial. is it is a polynomial, state the degree of the polynomial.

Svetlanka [38]

Yes, this is a polynomial. It is an expression containing variables raised to a real number.

In this case, the variables are being raised to the power of one.

The degree of this polynomial, thus, is one.

5 0

2 years ago

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PLEASE HELP!! NO LINKS!! GIVING 20 POINTS AWAY!!

Kruka [31]

Answer:

m∠N = 32°

NQ = 106°

When finding inscribed angles like ∠N with the intercepted arc, the equation is ∠N=1/2MP. (Inscribed angles are always half the degree of the arc length.) Plug in the corresponding value to get ∠N=1/2(64) to get 32°. When finding the angle of the intercepted arc with inscribed angles like NQ, the equation is NQ=2(∠P). Plug in the corresponding value to get 2(53) to get 106°.

5 0

2 years ago

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