Please help!!!!! One third of a number x is no more than 15.

Mathematics

1 answer:

Nitella [24]3 years ago

7 0

Answer:

1/3x is less than or equal to 15

Step-by-step explanation:

assuming you want an equation?

1/3x is less than or equal to 15

its this sign < with line under (srry i don't have the symbol)

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Y-3=-3y-43 solve the equation

Sonja [21]

Y-3=-3y-43
4y-3=-43
4y=-46
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8 0

3 years ago

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Samantha has 54 beads. Amanda has twice as many beads as Samantha. They are making bracelets it takes 8 beads to make a bracelet

wlad13 [49]

Answer:

On combining the beads they can make 20 bracelets.

Step-by-step explanation:

The number of beads Samantha has = 54 beads

Number of beads Amanda has = 2 x( Number of beads Samantha has)

= 2 x 54 = 108 beads

The total number of beads Amanda and Samantha has

= 54 beads + 108 beads

= 162 beads

Number of beads required to make 1 bracelet = 8 beads

Now, $\textrm{The number of bracelets made} = \frac{\textrm{Total number of beads they both have}}{\textrm{Beads required to make 1 bracelet}}$

= $\frac{162}{8} = 20.25$ ≈ 20

or, the total number of bracelets they can make is 20.

Hence, on combining the beads they can make 20 bracelets .

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3 years ago

Criminal investigators use biometric matching for fingerprint recognition, facial recognition, and iris recognition. When matchi

Art [367]

Answer:

It is not a Type I error neither a Type II error.

Step-by-step explanation:

Let $\mu$ be the true mean match score. The null hypothesis is $H_{0}: \mu = 80$ and the alternative hypothesis is $H_{1}: \mu > 80$ (upper-tail alternative). When the test shows that the mean match score is more than 80 when actually is equal to 80 a Type I error is made. On the other hand, when the test shows that the mean match score is equal to 80 when actually is more than 80 a type II error is made. Therefore, when the test shows that the mean match score is more than 80 when the person does not actually have a fingerprint match, does not correspond to a Type I error neither to a Type II error.