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2 years ago

Please help me I’ve asked this questions several times and I really need help with this I do not understand it at all

Mathematics

1 answer:

Ulleksa [173]2 years ago

6 0

Answer:

(-5,-2)

(-4,-3)

(-6,-1)

Step-by-step explanation:

-5(x)-2(y)=-7

-4(x)-3(y)=-7

-6(x)-1(y)=-7

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9x^2-1-(3x-2)^2=0 find the roots

Marizza181 [45]

Step-by-step explanation:

Make me as brain liest plz i need it

3 0

2 years ago

Simplify. Re-write in radical form last, simplify any roots if possible. (x^6)^1/8

GREYUIT [131]

I hope this helps you

x^6.1/8

x^3/4

4Vx^3

(4.square root of x^3)

3 0

3 years ago

The scale of the model train is 1 inches to 14.5 feet. What is the cars of the model train is 4 inches long. What is the length,

Len [333]

Pretty sure its 58, because i think you find it by multiplying the 4 inches by 14.5 because that's what each inch represents, so if this is wrong I'm sorry, but I'm pretty sure its 58.

8 0

2 years ago

Boyd’s dartboard is octagonal. Each side measures 9 in. He drew a copy of the board that was smaller by a scale factor 0.25.

Basile [38]

The question is about the side length of a copy of the dartboard. We know that the original dartboard has sides that measure 9 inches. We also have the scale factor 0.25. The ratio of the corresponding sides is called the scale factor. If we have the scale factor 0.25 it means that for 1 unit of the original figure there are 0.25 units of length of a copy. Therefore: x = 9 * 0.25; x = 2.25 in. Answer: The side length of the dartboard Boyd drew is 2.25 inches.

5 0

3 years ago

The time for this event for boys in secondary school is known to possess a normal distribution with a mean of 460 seconds and a

AlekseyPX

Answer:

The time that the boys need to beat in order to earn a certificate of recognition from the fitness association is 511.264 seconds.

Step-by-step explanation:

We are given that the time for this event for boys in secondary school is known to possess a normal distribution with a mean of 460 seconds and a standard deviation of 40 seconds.

The fitness association wants to recognize the fastest 10% of the boys with certificates of recognition.

Let X = time for this event for boys in secondary school

SO, X ~ Normal( $\mu=460,\sigma^{2} =40^{2}$ )

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

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

where, $\mu$ = mean time = 460 seconds

$\sigma$ = standard deviation = 40 seconds

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.

Now, it is given that the fitness association wants to recognize the fastest 10% of the boys with certificates of recognition, which means;

P(X > x) = 0.10 {where x is the required time which boy need to beat}

P( $\frac{X-\mu}{\sigma}$ > $\frac{x-460}{40}$ ) = 0.10