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nadya68 [22]
3 years ago
12

What is the truth value for the following conditional statement?

Mathematics
2 answers:
alexgriva [62]3 years ago
8 0
True I think is the answer
Zinaida [17]3 years ago
5 0

Answer:  the correct option is

(B) F T → T.

Step-by-step explanation:  We are given to find the truth value of he following conditional statement :

"If the grass is red, then the snow is white."

Let us consider that

p : the grass is red. This is a false statement

and  

q : the snow is white. This is a true statement.

So, we get

p  →  q

F  → T, which is true (T).

Thus, the required answer is  F T → T

Option (B) is CORRECT.

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I dont get this it's geometry.
SIZIF [17.4K]
Angle 2 is half of 115 so it would be 57.5 and angle 1 would be 122.5 degrees
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3 years ago
Choose the equation that shows a step in the process of completing the square on the given quadratic. y = x^2 + 8x – 3
klasskru [66]

Answer:

y = x^2 + 8x - 3

y = x^2 + 8x + (8/2)^2 - 3 - (8/2)^2

y = x^2 + 8x + 16 - 3 - 16

Step-by-step explanation:

4 0
3 years ago
Given that f(x) = x2 + 12x + 35 and g(x) = x + 7, find (f:g)(x) and express
rusak2 [61]

Answer:

2

Step-by-step explanation:

4 0
3 years ago
The researchers are concerned that the dollars spent per shopper examined in the study is too different from what has been found
Katarina [22]

Answer:

Step-by-step explanation:

Hello!

To study if the average dollars spend per shopper is different than the expected population average of $84.5 there was a sample of 120 shoppers taken and their spending habit registered.

Then the study variable is:

X: Amount of money spent by a shopper ($)

This population standard deviation is σ= $16

Using the data information I've run a normality test, with p-value: 0.3056 against the test significance level α: 0.05, the decision is to not reject the null hypothesis so there is enough statistical evidence to conclude that the study variable has a normal distribution.

X~N(μ;σ²)

I've also calculated the sample mean and standard deviation:

X[bar]= $81.01

S= $19.62

a.

If the objective is to test that the true mean spending for the population is significantly different than the expected average of 84.5, the parameter of interest is the population mean μ and the hypothesis test will be two-tailed.

H₀: μ = 84.5

H₁: μ ≠ 84.5

α:0.05

The statistic is a standard normal for one sample:

Z= \frac{X[bar]-Mu}{\frac{Sigma}{\sqrt{n} } } ~N(0;1)

Z_{H_0}= \frac{81.01-84.5}{\frac{16}{\sqrt{120} } } = -2.389

p-value two tailed  0.016894

b.

The decision rule using the p-value method is:

If p-value ≤ α ⇒ Reject the null hypothesis.

If p-value > α ⇒ Do not reject the null hypothesis.

Since the p-value is less than the significance level, the decision is to reject the null hypothesis.

c.

Using a level of significance of 5% the null hypothesis was rejected. So there is enough statistical evidence to support the researcher's claim that the true mean spending for the population is significantly different than the expected average of $84.5.

I hope it helps!

3 0
3 years ago
If y varies with x, and y=-20 and x=5, what is x when y=-30?
vladimir1956 [14]

If y = -30, then x = 7.5

Hope this helps.

3 0
3 years ago
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