All categories

2 years ago

Plot -1/4 on the horizontal and vertical number lines.

Mathematics

1 answer:

damaskus [11]2 years ago

8 0

On the vertical number line, negative numbers are to the bottom of 0. Therefore, -1/4 will go one unit down from 0.

On the horizontal number line, negative numbers are to the left of 0. Therefore, -1/4 will go to the left one unit.

Best of Luck!

You might be interested in

Evaluate the expression: 0.5f - 2.3g where f = 12 and g = 2

m_a_m_a [10]

Answer:

1.4

Step-by-step explanation:

0.5f-2.3g where f=12 and g=2

substitute 12 in for f and 2 in for g and solve

0.5(12)-2.3(2)

6-4.6=1.4

4 0

2 years ago

Read 2 more answers

Need help ASAP!!!!!!

trapecia [35]

Answer:

see explanation

Step-by-step explanation:

x² + 3x + 7 = 5 ( subtract 5 from both sides )

x² + 3x + 2 = 0 ← in standard form

(x + 2)(x + 1) = 0 ← in factored form

Equate each factor to zero and solve for x ( zero product rule )

x + 2 = 0 → x = - 2

x + 1 = 0 ⇒ x = - 1

--------------------------------------------------------------

x² - 2 = - 2x² + 5x ( subtract - 2x² + 5x from both sides )

3x² - 5x - 2 = 0 ← in standard form

(3x + 1)(x - 2) = 0 ← in factored form

Equate each factor to zero and solve for x

3x + 1 = 0 ⇒ 3x = - 1 ⇒ x = - $\frac{1}{3}$

x - 2 = 0 ⇒ x = 2

------------------------------------------------------------

(x + 3)² + 4x = 0 ← expand left side using FOIL and simplify

x² + 6x + 9 + 4x = 0

x² + 10x + 9 = 0 ← in standard form

(x + 9)(x + 1) = 0 ← in factored form

Equate each factor to zero and solve for x

x + 9 = 0 ⇒ x = - 9

x + 1 = 0 ⇒ x = - 1

4 0

2 years ago

The weights of adult male birds of a certain species are normally distributed with a mean of 27.5 grams and a standard deviation

Korvikt [17]

Answer:

The z-score of a male bird of this species with a weight of 29.37 grams is 1.7.

Step-by-step explanation:

We are given that the weights of adult male birds of a certain species are normally distributed with a mean of 27.5 grams and a standard deviation of 1.1 grams.

Let X = weights of adult male birds of a certain species

So, X ~ Normal( $\mu=27.5,\sigma^{2} =1.1^{2}$ )

The z score probability distribution for normal distribution is given by;

Z = $\frac{ X-\mu}{\sigma } }$ ~ N(0,1)

where, $\mu$ = population mean weight = 27.5 grams

$\sigma$ = standard deviation = 1.1 grams

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

SO, the z-score of a male bird of this species with a weight of 29.37 grams is given by;

Z score = $\frac{ X-\mu}{\sigma } }$ = $\frac{ 29.37-27.5}{1.1 } }$