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2 years ago

SOMEONE HELP PLEASE

Mathematics

1 answer:

Alborosie2 years ago

4 0

I’m sorry but I can’t understand what the question is saying

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What is the value of x in the equation 2x + 3y = 36, when y = 6? 8 9 27 36

Nimfa-mama [501]

Answer:

x = 9

Step-by-step explanation:

we can plug in the given value for y into the equation to find x

2x + 3y = 36

2x + 3(6) = 36

2x + 18 = 36

2x = 18

x = 9

5 0

2 years ago

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Estimate the product of 17 and 234 4,000 2,000 6,000 3,000

zubka84 [21]

Answer:

I believe the answer is D?

Step-by-step explanation:

8 0

3 years ago

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Suppose babies born after a gestation period of 32 to 35 weeks have a mean weight of 2,700 grams and a standard deviation of 900

makkiz [27]

Answer:

the answers have been provided below

Step-by-step explanation:

Population z score is given as (x-μ)/sd For a 32 to 35 week gestation period

μ= 2700

Sd = 900grams

x = 2225

z = (2225-2700)/900

z = -0.527

For a 40 week gestation period.

μ = 3000

Sd = 465grams

x = 2525

z = (2525-3000)/465

Z = -1.022

In answer to the question:

a. The 33-week gestation period baby weighs 0.53 standard deviations below the mean.

b. The 40-week gestation period baby weighs 1.02 standard deviations below the mean.

c. -1.022 < -0.527. the 40 weeks gestation baby weighs less.

A. The baby born in week 40 does since its z-score is smaller.

6 0

3 years ago

A bag of 100 tulip bulbs purchased from a nursery contains 35 red tulip bulbs, 35 yellow tulip bulbs, and 30 purple tulip bulb

Andrej [43]

A) 35/100= 35%
B)30/100= 30%

5 0

3 years ago

The angle \theta_1θ 1 theta, start subscript, 1, end subscript is located in Quadrant \text{II}IIstart text, I, I, end text, a

andrew11 [14]

Answer:

$\cos(\theta_1)=-\dfrac{\sqrt{15}}{4}$

Step-by-step explanation:

The angle $\theta_1$ is located in Quadrant II.

$\sin(\theta_1)=\dfrac{1}{4}$

From trigonometry, we know that:

$\sin(\theta)=\dfrac{Opposite}{Hypotenuse}\\$Therefore:\\Opposite=1\\Hypotenuse=4\\Using Pythagorean theorem:\\Hypotenuse^2=Opposite^2+Adjacent^2\\4^2=1^2+Adjacent^2\\Adjacent^2=16-1\\Adjacent^2=15\\Adjacent=\sqrt{15}$

Now, in Quadrant II,

The x-axis is negative
The y-axis is positive

Therefore, the Adjacent angle to $\theta_1 =-\sqrt{15}$

Therefore:

$\cos(\theta_1)=\dfrac{Adjacent}{Hypotenuse}=\dfrac{-\sqrt{15}}{4}\\\\\cos(\theta_1)=-\dfrac{\sqrt{15}}{4}$

8 0

3 years ago