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2 years ago

The mean temperature for the first 7 days in January was 4 °C.

1 answer:

6 0

The mean temperature would be 5 c.
You figure this out by adding both values, 4 and 6, and dividing it by 2 because there are 2 values. So the equation would be 4 + 6 = 10
10 / 2 = 5

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What are the solutions to the quadratic equation below?

Luden [163]

Using factorization method;

x²+13x+40=0

x²+13+40

Find two numbers that you can add together to give the co-efficient of variable x. (I.e 13 for this question). Also, you'll find two numbers that you can multiply with each other to give you the whole number as an answer. (I.e 40 in this question). The two numbers must be the same (i.e the two numbers that will be added to give the co-efficient of x and the two numbers that will be multiplied to give the whole number must be the same two).

The two numbers are +5 and +8

The equation will therefore be = x²+5x+8x+40

You'll then factorize (I.e use a common factor of both values to bracket them)

x(x+5)+8(x+5)

(x+8)(x+5)

x is therefore (x+8=0) or (x+5=0)

x=(x=0-8) or (x=0-5)

x= -8 or x= -5

5 0

1 year ago

Please help me! simplify. 5-[-(-20] a. -3 b.3 c.-7 d.7

frosja888 [35]

Im sorry i dont know

8 0

3 years ago

Simplify negative 5 minus the square root of negative 44 negative 5 minus 4 times the square root of 11 i negative 5 minus 4 i t

Anika [276]

Answer:

$-5-\sqrt{-44}=-5-2\sqrt{11}i$

Step-by-step explanation:

We want to simplify:

$-5-\sqrt{-44}$

We rewrite the expression under the radical sign to obtain:

$-5-\sqrt{4\times11\times -1}$

We split the expression under the radical sign to get:

$-5-\sqrt{4} \times \sqrt{11} \times \sqrt{-1}$

Recall that:

$\sqrt{-1}=i$

This implies that:

$-5-\sqrt{4} \times \sqrt{11} \times \sqrt{-1}=-5-2\sqrt{11}i$

Therefore $-5-\sqrt{-44}=-5-2\sqrt{11}i$

6 0

3 years ago

Which equation is not equal to 6^3/6^6 ?: 1/6^2 6^-3 1/216 1/6^3

SCORPION-xisa [38]

The answer would be 60 because its multiplication all those numbers

4 0

2 years ago

In a standard normal distribution, what z value corresponds to 17% of the data between the mean and the z value?

hjlf

You're looking for a value $z$ such that

$\mathbb P(0$

Because the distribution is symmetric, the value of $z$ in either case will be the same.

Now, because the distribution is continuous, you have that

$0.17=\mathbb P(0$

The mean for the standard normal distribution is $0$ , and because the distribution is symmetric about its mean, it follows that $\mathbb P(Z$ .

$0.17=\mathbb P(Z$

You can consult a $z$ score table to find the corresponding score for this probability. It turns out to be $z\approx0.4399$ .

4 0

2 years ago