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3 years ago

12

The teacher of a jewelry making class had a supply of 236 beads.Her students used 29 beads to make earrings and 63 beads to make

bracelets.They will use the.They will use the remaining beads to make necklaces with 6 beads on each necklace?How many necklaces will the students make

1 answer:

AysviL [449]3 years ago

4 0

$236 - 29 - 63 = 144 \\ 144 \div 6 = 24 \\$
24 necklaces would be made

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PLEASE HELP 8th GRADE MATH QUESTION OVER HERE!!!

patriot [66]

The side AB measures option 2. $\sqrt{20}}$ units long.

Step-by-step explanation:

Step 1:

The coordinates of the given triangle ABC are A (4, 5), B (2, 1), and C (4, 1).

The sides of the triangle are AB, BC, and CA. We need to determine the length of AB.

To calculate the distance between two points, we use the formula $d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}.$

where ( $x_{1},y_{1}$ ) are the coordinates of the first point and ( $x_{2},y_{2}$ ) are the coordinates of the second point.

Step 2:

For A (4, 5) and B (2, 1), ( $x_{1},y_{1}$ ) = (4, 5) and ( $x_{2},y_{2}$ ) = (2, 1). Substituting these values in the distance formula, we get

$d=\sqrt{\left(2-4\right)^{2}+\left(1-5}\right)^{2}} = \sqrt{\left(2\right)^{2}+\left(4}\right)^{2}}=\sqrt{20}}.$

So the side AB measures $\sqrt{20}}$ units long which is the second option.

6 0

3 years ago

The hire purchase cost of a radio is $6355. If the deposit is $1000 and the balance is to be paid in 9 equal instalments, each i

Zepler [3.9K]

Answer: $595

is the correct answer

4 0

3 years ago

Ms.Lopez has created a floor plan of a dollhouse. The area of the entire dollhouse is 576 square inches

harina [27]

Answer:

(a)Area of the bedroom=96 square inches

(b)Area of the living room =144 square inches

(c)Area of the dollhouse that is not of the bedroom or living room =336 square inches.

Step-by-step explanation:

The area of the bedroom is 1/6 the area of the entire dollhouse

The area of the living room is 3/2 times the area of the bedroom.

The area of the entire dollhouse is 576 square inches

Let the area of bedroom=b

Let the area of living room=l

(a) The area of the bedroom is 1/6 the area of the entire dollhouse

$b=\frac{1}{6}X576=96$ square inches

(b)The area, in square inches, of the living room

The area of the living room is 3/2 times the area of the bedroom

l= $\frac{3}{2}X96=144$ square inches

(c)The total area of the house =576 square inches

Area of the bedroom=96 square inches

Area of the living room =144 square inches

Area of the dollhouse that is not of the bedroom or living room

=576-(96+144)=576-240=336 square inches.

The area of the living room is 3/2 times the area of the bedroom

This is gotten by subtracting the area of the bedroom and living room from the total area of the house.

3 0

4 years ago

What is the answer for number 7?

aivan3 [116]

Answer:

Step-by-step explanation:

x=30

m=abe=180

8 0

3 years ago

Suppose that from the past experience a professor knows that the test score of a student taking his final examination is a rando

DENIUS [597]

Answer:

$n=13.167^2 =173.369$ and if we round up to the nearest integer we got n =174

Step-by-step explanation:

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Let X the random variable who represents the test score of a student taking his final examination. We know from the problem that the distribution for the random variable X is given by:

$X\sim N(\mu =73,\sigma =10.5)$

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

Solution to the problem

We want to find the value of n that satisfy this condition:

$P(71.5 < \bar X$

And we can use the z score formula given by:

$z=\frac{\bar X- \mu}{\frac{\sigma}{\sqrt{n}}}$

And we have this:

$P(\frac{71.5-73}{\frac{10.5}{\sqrt{n}}} < Z$

And we can express this like this:

$P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=0.94$

And by properties of the normal distribution we can express this like this:

$P(-0.14286 \sqrt{n} < Z< 0.14286 \sqrt{n} )=1-2P(Z$

If we solve for $P(Z$ we got:

$P(Z$

Now we can find a quantile on the normal standard distribution that accumulates 0.03 of the area on the left tail and this value is: $z=-1.881$

And using this we have this equality:

$-1.881 = -0.14286 \sqrt{n}$

If we solve for $\sqrt{n}$ we got:

$\sqrt{n} = \frac{-1.881}{-0.14286}=13.167$

And then $n=13.167^2 =173.369$ and if we round up to the nearest integer we got n =174

6 0

3 years ago