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2 years ago

PLZ HELP ASAP!! ILL GIVE BRAINLEST

Mathematics

2 answers:

Ira Lisetskai [31]2 years ago

5 0

Answer:

Just follow what it says

the red dot is the answer

Step-by-step explanation:

11111nata11111 [884]2 years ago

4 0

Answer:

(-5,-4)

Step-by-step explanation:

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Richard receives a paycheck twice each month. Here’s a copy of one of his pay statements. If there are 2 paydays per month, what

iogann1982 [59]

For the answer to the question above asking what is Richard’s monthly net income if Richard receives a paycheck twice each month. Here’s a copy of one of his pay statements. If there are 2 paydays per month?
The answer for this is first solved for his Net Pay then you can multiply it by two so that you can get his monthly net income.

3 0

3 years ago

GreenBeam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a m

Harrizon [31]

Answer:

(a) Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg

(b) The value of z test statistics is 2.50.

(c) We conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

(d) The p-value is 0.0062.

Step-by-step explanation:

We are given that Green Beam Ltd. claims that its compact fluorescent bulbs average no more than 3.50 mg of mercury. A sample of 25 bulbs shows a mean of 3.59 mg of mercury. Assuming a known standard deviation of 0.18 mg.

Let $\mu$ = average mg of mercury in compact fluorescent bulbs.

So, Null Hypothesis, $H_0$ : $\mu \leq$ 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is no more than 3.50 mg}

Alternate Hypothesis, $H_A$ : $\mu$ > 3.50 mg {means that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg}

The test statistics that would be used here One-sample z test statistics as we know about the population standard deviation;

T.S. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean mg of mercury = 3.59

$\sigma$ = population standard deviation = 0.18 mg

n = sample of bulbs = 25

So, test statistics = $\frac{3.59-3.50}{\frac{0.18}{\sqrt{25} } }$

= 2.50

The value of z test statistics is 2.50.

Now, P-value of the test statistics is given by;

P-value = P(Z > 2.50) = 1 - P(Z $\leq$ 2.50)

= 1 - 0.9938 = 0.0062

Now, at 0.01 significance level the z table gives critical value of 2.3263 for right-tailed test. Since our test statistics is more than the critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the average mg of mercury in compact fluorescent bulbs is more than 3.50 mg.

6 0

3 years ago

What are the ratios of sine, cosine, and tangent for angle Y? sin(Y) = StartFraction X Z Over X Y EndFraction; cos(Y) = StartFra

Sholpan [36]

Answer:

$\sin Y= \frac{XZ}{XY}$

$\cos Y= \frac{YZ}{XY}$

$\tan Y= \frac{XZ}{YZ}$

Step-by-step explanation:

Given

See attachment for triangle

Required

Find $\sin, \cos$ and $\tan$ of angle Y

For angle Y:

$Opposite = XZ$

$Adjacent = YZ$

$Hypotenuse = XY$

The $\sin$ of an angle is calculated as:

$\sin\theta = \frac{Opposite}{Hypotenuse}$

So:

$\sin Y= \frac{XZ}{XY}$

The $\cos$ of an angle is calculated as:

$\cos\theta = \frac{Adjacent}{Hypotenuse}$

So:

$\cos Y= \frac{YZ}{XY}$

The $\tan$ of an angle is calculated as:

$\tan\theta = \frac{Opposite}{Adjacent}$

So:

$\tan Y= \frac{XZ}{YZ}$

7 0

2 years ago

Read 2 more answers

There is a triangle with a perimeter of 63 cm, one side of which is 21 cm. Also, one of the medians is perpendicular to one of t

NemiM [27]

Answer:

21cm; 28cm; 14cm

Step-by-step explanation:

There is no info in the problem/s text which one of the triangle's side is 21 cm. That is why we have to try all possible variants.

Let the triangle is ABC . Let the AK is the angle A bisector and BM is median.

Let O is AK and BM cross point.

Have a look to triangle ABM. AO is the bisector and AOB=AOM=90 degrees (means that AO is as bisector as altitude)

=> triangle ABM is isosceles => AB=AM (1)

1. Let AC=21 So AM=21/2=10.5 cm

So AB=10.5 cm as well. So BC= P-AB-AC=63-21-10.5=31.5 cm

Such triangle doesn' t exist ( is impossible), because the triangle's inequality doesn't fulfill. AB+AC>BC ( We have 21+10.5=31.5 => AB+AC=BC)

2. Let AB=21 So AM=21 and AC=42 .So BC= P-AB-AC=63-21-42=0 cm- such triangle doesn't exist.

3. Finally let BC=21 cm. So AB+AC= 63-21=42 cm

We know (1) that AB=AM so AC=2*AB. So AB+AC=AB+2*AB=3*AB

=>3*AB=42=> AB=14 cm => AC=2*14=28 cm.

Let check if this triangle exists ( if the triangle's inequality fulfills).

BC+AB>AC 21+14>28 - correct=> the triangle with the sides' length 21cm,14 cm, 28cm exists.

This variant is the only possible solution of the given problem.

6 0

3 years ago

Using the given postulate, tell which parts of the pair of triangles should be shown congruent.

weqwewe [10]

Answer:

it should be the 5th one because it shows defencicy

Step-by-step explanation:

from what i know

4 0

2 years ago