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3 years ago

Can someone please help me i will mark brainliest

Mathematics

2 answers:

docker41 [41]3 years ago

6 0

Slope=7/2=3.5
Up 7, over 2

alekssr [168]3 years ago

6 0

Slope = 7/5 = 3.5
Up 7, over 2
That’s should be your answer
Hopefully I’m right :)

You might be interested in

In circle N with m See diagram below

Alex777 [14]

Answer:

$113.10$

Step-by-step explanation:

The area of a sector with measure $\theta$ and radius $r$ is given by $A_{sec}=r^2\pi\cdot \frac{\theta}{360^{\circ}}$ .

What we're given:

$r$ of 12
$\theta$ of $90^{\circ}$

Substituting given values, we get:

$A_{sec}=12^2\pi\cdot \frac{90}{360},\\\\A_{sec}=144\pi\cdot \frac{1}{4},\\\\A_{sec}\approx \boxed{113.10}$

8 0

3 years ago

Activity 5: COMPARE US! Write an inequality relating the given pair of angles or segment measures. Complete each statement by ch

Shalnov [3]

Answer:

1. LM < PN

2. AD < DC

3. m<CAB < m<CBA

4. m<1 = m<2

Step-by-step explanation:

Recall: an angle measure is relative to the length of the opposite side. That is, the longer the side opposite to an angle, the larger the measure of that angle and vice versa.

1. LM is opposite to <LNM,

PN is opposite to <NLP

m<LNM is less than m<NLP, therefore,

LM < PN

2. AD is opposite to <ABD

DC is opposite to <DBC

m<ABD is less than m<DBC, therefore,

AD < DC

3. m<CAB is opposite to CB

m<CBA is opposite to CA

CB is less than CA, therefore,

m<CAB < m<CBA

4. The side opposite to <1 is congruent to the side opposite to <2.

Therefore,

m<1 = m<2

3 0

3 years ago

How to solve this algebraically?

Keith_Richards [23]

Ok so x+4-1 makes x-1 so i belive the answer could be 1/5

7 0

3 years ago

Read 2 more answers

A half-century ago, the mean height of women in a particular country in their 20s was 64.7 inches. Assume that the heights of

Ainat [17]

Answer:

99.5% of all samples of 21 of today's women in their 20's have mean heights of at least 65.86 inches.

Step-by-step explanation:

We are given that a half-century ago, the mean height of women in a particular country in their 20's was 64.7 inches. Assume that the heights of today's women in their 20's are approximately normally distributed with a standard deviation of 2.07 inches.

Also, a samples of 21 of today's women in their 20's have been taken.

Let $\bar X$ = sample mean heights

The z-score probability distribution for sample mean is given by;

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

where, $\mu$ = population mean height of women = 64.7 inches

$\sigma$ = standard deviation = 2.07 inches

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, Probability that the sample of 21 of today's women in their 20's have mean heights of at least 65.86 inches is given by = P( $\bar X$ $\geq$ 65.86 inches)

P( $\bar X$ $\geq$ 65.86 inches) = P( $\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n} } }$ $\geq$ $\frac{65.86-64.7}{\frac{2.07}{\sqrt{21} } }$ ) = P(Z $\geq$ -2.57) = P(Z $\leq$ 2.57)

= 0.99492 or 99.5%

The above probability is calculated by looking at the value of x = 2.57 in the z table which has an area of 0.99492.

Therefore, 99.5% of all samples of 21 of today's women in their 20's have mean heights of at least 65.86 inches.

3 0

3 years ago

Need help on number one and three

Naya [18.7K]

An example for #1 would be that: 15 is divisible by 3, but not 9. every third multiple of 3, (9, 18, 27,...) is divisible by 9 because 9 is three times the size of 3.

For part two the first number that I thought of is 23.

4 0

3 years ago